Let us find the volume of the body generated by the rotation of the cycloid arch around its base. Roberval found it by breaking the resulting egg-shaped body (Fig. 5.1) into infinitely thin layers, inscribing cylinders into these layers and adding up their volumes. The proof turned out to be long, tedious and not entirely rigorous. Therefore, to calculate it, we turn to higher mathematics. Let us define the equation of the cycloid parametrically.
In integral calculus, when studying volumes, the following remark is used:
If the curve bounding a curvilinear trapezoid is given by parametric equations and the functions in these equations satisfy the conditions of the change of variable theorem in a definite integral, then the volume bodies of revolution trapezoid around the Ox axis, will be calculated by the formula:
Let's use this formula to find the volume we need.
In the same way, we calculate the surface of this body.
L=((x,y): x=a(t - sin t), y=a(1 - cost), 0 ? t ? 2р)
In integral calculus, there is the following formula for finding the surface area of a body of rotation around the x-axis of a curve defined parametrically on a segment (t 0 ?t ?t 1):
Applying this formula to our cycloid equation we get:
Let us also consider another surface generated by the rotation of the cycloid arc. To do this, we will construct a mirror image of the cycloid arch relative to its base, and we will rotate the oval figure formed by the cycloid and its reflection around the KT axis (Fig. 5.2)
First, let's find the volume of the body formed by the rotation of the cycloid arch around the KT axis. We will calculate its volume using the formula (*):
Thus, we calculated the volume of half of this turnip-shaped body. Then the entire volume will be equal
Before moving on to the formulas for the area of a surface of revolution, we will give a brief formulation of the surface of revolution itself. A surface of revolution, or, what is the same thing, a surface of a body of revolution is a spatial figure formed by the rotation of a segment AB curve around the axis Ox(picture below).
Let us imagine a curved trapezoid bounded from above by the mentioned segment of the curve. A body formed by rotating this trapezoid around the same axis Ox, and is a body of rotation. And the area of the surface of revolution or the surface of a body of revolution is its outer shell, not counting the circles formed by rotation around the axis of straight lines x = a And x = b .
Note that a body of revolution and, accordingly, its surface can also be formed by rotating the figure not around the axis Ox, and around the axis Oy.
Calculating the area of a surface of revolution specified in rectangular coordinates
Let in rectangular coordinates on the plane the equation y = f(x) a curve is specified, the rotation of which around the coordinate axis forms a body of revolution.
The formula for calculating the surface area of revolution is as follows:
(1).
Example 1. Find the surface area of the paraboloid formed by rotation around its axis Ox arc of a parabola corresponding to the change x from x= 0 to x = a .
Solution. Let us express explicitly the function that defines the arc of the parabola:
Let's find the derivative of this function:
Before using the formula to find the area of a surface of revolution, let’s write that part of its integrand that represents the root and substitute the derivative we just found there:
Answer: The length of the arc of the curve is
.
Example 2. Find the surface area formed by rotation around an axis Ox astroid.
Solution. It is enough to calculate the surface area resulting from the rotation of one branch of the astroid, located in the first quarter, and multiply it by 2. From the astroid equation, we will explicitly express the function that we will need to substitute into the formula to find the surface area of rotation:
.
We integrate from 0 to a:
Calculation of the area of a surface of revolution specified parametrically
Let us consider the case when the curve forming the surface of revolution is given by parametric equations
Then the surface area of rotation is calculated by the formula
(2).
Example 3. Find the area of the surface of revolution formed by rotation around an axis Oy figure bounded by a cycloid and a straight line y = a. The cycloid is given by parametric equations
Solution. Let's find the intersection points of the cycloid and the straight line. Equating the equation of a cycloid and the equation of a straight line y = a, let's find
It follows from this that the boundaries of integration correspond to
Now we can apply formula (2). Let's find derivatives:
Let's write the radical expression in the formula, substituting the found derivatives:
Let's find the root of this expression:
.
Let's substitute what we found into formula (2):
.
Let's make a substitution:
And finally we find
Trigonometric formulas were used to transform expressions
Answer: The surface area of revolution is .
Calculating the area of a surface of revolution specified in polar coordinates
Let the curve, the rotation of which forms the surface, be specified in polar coordinates.
Lectures 8. Applications of a definite integral.
The application of the integral to physical problems is based on the property of the additivity of the integral over a set. Therefore, using the integral, quantities can be calculated that are themselves additive in the set. For example, the area of a figure is equal to the sum of the areas of its parts. The length of the arc, surface area, volume of the body, and mass of the body have the same property. Therefore, all these quantities can be calculated using a definite integral.
You can use two methods to solve problems: method of integral sums and method of differentials.
The method of integral sums repeats the construction of a definite integral: a partition is constructed, points are marked, the function is calculated at them, the integral sum is calculated, and passage to the limit is performed. In this method, the main difficulty is to prove that in the limit the result is exactly what is needed in the problem.
The differential method uses the indefinite integral and the Newton–Leibniz formula. The differential of the quantity to be determined is calculated, and then, by integrating this differential, the required quantity is obtained using the Newton–Leibniz formula. In this method, the main difficulty is to prove that it is the differential of the required value that has been calculated, and not something else.
Calculation of areas flat figures.
1. The figure is limited by the graph of a function defined in a Cartesian coordinate system.
We came to the concept of a definite integral from the problem of the area of a curved trapezoid (in fact, using the method of integral sums). If the function only accepts negative values, then the area under the graph of a function on a segment can be calculated using a definite integral. notice, that 
therefore, the method of differentials can also be seen here.
But a function can also take negative values on a certain segment, then the integral over this segment will give a negative area, which contradicts the definition of area.
You can calculate the area using the formulaS=. This is equivalent to changing the sign of the function in those areas in which it takes negative values.
If you need to calculate the area of a figure bounded above by the graph of the function and below by the graph of the function, then you can use the formulaS=
, because .
Example. Calculate the area of the figure bounded by straight lines x=0, x=2 and graphs of functions y=x 2, y=x 3.
Note that on the interval (0,1) the inequality x 2 > x 3 holds, and for x >1 the inequality x 3 > x 2 holds. That's why
2. The figure is limited by the graph of a function specified in a polar coordinate system.
Let the graph of a function be given in a polar coordinate system and we want to calculate the area of a curvilinear sector bounded by two rays and the graph of a function in a polar coordinate system.
Here you can use the method of integral sums, calculating the area of a curvilinear sector as the limit of the sum of the areas of elementary sectors in which the graph of the function is replaced by a circular arc 
.
You can also use the differential method: 
.
You can think like this. Replacing the elementary curvilinear sector corresponding to the central angle with a circular sector, we have the proportion . From here 
. Integrating and using the Newton–Leibniz formula, we obtain 
.
Example. Let's calculate the area of the circle (check the formula). We believe. The area of the circle is 
.
Example. Let's calculate the area bounded by the cardioid 
.
3 The figure is limited by the graph of a function specified parametrically.
The function can be specified parametrically in the form . We use the formula S=
, substituting into it the limits of integration over the new variable. 
. Usually, when calculating the integral, those areas where the integrand function has a certain sign are isolated and the corresponding area with one or another sign is taken into account.
Example. Calculate the area enclosed by the ellipse.
Using the symmetry of the ellipse, we calculate the area of the quarter of the ellipse located in the first quadrant. In this quadrant. That's why .
Calculation of volumes of bodies.
1. Calculation of volumes of bodies from the areas of parallel sections.
Let it be required to calculate the volume of a certain body V from the known cross-sectional areas of this body by planes perpendicular to the line OX drawn through any point x of the line segment OX.
Let's apply the method of differentials. Considering the elementary volume, over a segment the volume of a straight line circular cylinder with base area and height, we get 
. Integrating and applying the Newton–Leibniz formula, we obtain
2. Calculation of volumes of bodies of rotation.
Let it be necessary to calculate OX.
Then 
.
Likewise, volume of a body of revolution around an axisOY, if the function is given in the form , can be calculated using the formula .
If the function is specified in the form and it is required to determine the volume of a body of rotation around an axisOY, then the formula for calculating the volume can be obtained as follows.
Passing to the differential and neglecting the quadratic terms, we have 
. Integrating and applying the Newton–Leibniz formula, we have .
Example. Calculate the volume of the sphere.
Example. Calculate the volume of a right circular cone bounded by a surface and a plane.
Let's calculate the volume as the volume of a body of rotation formed by rotation around the OZ axis right triangle in the OXZ plane, the legs of which lie on the OZ axis and the line z = H, and the hypotenuse lies on the line.
Expressing x in terms of z, we get 
.
Calculation of arc length.
In order to obtain formulas for calculating the length of an arc, recall the formulas derived in the 1st semester for the differential of the arc length.
If the arc is the graph of a continuously differentiable function, the arc length differential can be calculated using the formula
. That's why 
If a smooth arc is specified parametrically, That
. That's why 
.
If the arc is specified in a polar coordinate system, That
. That's why 
.
Example. Calculate the length of the arc of the graph of the function, . 
.
In lessons about equation of a straight line on a plane And equations of a straight line in space.
Meet an old friend:
The curvilinear trapezoid is proudly crowned with a graph, and, as you know, it area is calculated using a definite integral according to the elementary formula or, in short: .
Let's consider the situation when same function specified in parametric form.
How to find the area in this case?
At some quite specific value of the parameter, the parametric equations will determine the coordinates of the point, and for another quite specific value – coordinates of the point. When “te” changes from to inclusive, the parametric equations “draw” the curve. I think everything has become clear about the limits of integration. Now into the integral instead of“X” and “Y” we substitute the functions and open the differential:
Note : it is assumed that the functions continuous on the integration interval and, in addition, the function monotonous On him.
The formula for the volume of a body of revolution is just as simple:
The volume of a body obtained by rotating a curved trapezoid around the axis is calculated by the formula 
or: . We substitute parametric functions into it, as well as integration limits:
Please record both working formulas in your reference book.
According to my observations, problems on finding volume are quite rare, and therefore a significant part of the examples in this lesson will be devoted to finding area. Let's not put things off for a long time:
Example 1
Calculate the area of a curved trapezoid 
, If
Solution: use the formula 
.
A classic problem on a topic that is understood always and everywhere:
Example 2
Calculate the area of an ellipse
Solution: for definiteness, we assume that the parametric equations define canonical ellipse with the center at the origin, semi-major axis “a” and semi-minor axis “be”. That is, according to the condition, we are offered nothing more than
find the area of the ellipse
It is obvious that parametric functions are periodic, and . It would seem that you can charge the formula, but not everything is so transparent. Let's find out direction, in which parametric equations “draw” an ellipse. As a guide, we will find several points that correspond to the simplest parameter values: 
It is easy to understand that when the parameter “te” changes from zero to “two pi”, the parametric equations “draw” an ellipse counterclock-wise:
Due to the symmetry of the figure, we calculate the part of the area in the 1st coordinate quarter, and multiply the result by 4. Here we see fundamentally the same picture that I commented just above: the parametric equations “draw” the arc of the ellipse “in the opposite direction” of the axis, but the area figures are counted from left to right! That's why lower the limit of integration corresponds to the value, and top limit – value .
As I already advised in the lesson Area in polar coordinates, quadruple the result is better At once:
The integral (if someone suddenly discovered such an incredible gap) was analyzed in class Integrals of trigonometric functions.
Answer:
Essentially, we have derived a formula for finding the area ellipse. And if in practice you come across a task with specific values of “a” and “be”, then you can easily perform a reconciliation/check, since the problem is solved in a general form.
The area of the ellipse is also calculated in rectangular coordinates; to do this, you need to express the “y” from the equation and solve the problem exactly as in Example No. 4 of the article Efficient methods for solving definite integrals. Be sure to look at this example and compare how much easier it is to calculate the area of an ellipse if it is defined parametrically.
And, of course, I almost forgot, parametric equations can define a circle or ellipse in a non-canonical position.
Example 3
Calculate the area of one arc of a cycloid 
To solve a problem, you need to know what it is cycloid or at least purely formally complete the drawing. A sample design at the end of the lesson. However, I won’t send you far away; you can look at the graph of this line in the following problem:
Example 4
Solution: parametric equations 
define a cycloid, and the constraint indicates the fact that we are talking about its first arch, which is “drawn” when the parameter value changes within . Please note that here is the “correct” direction of this “drawing” (from left to right), which means there will be no problems with the limits of integration. But a bunch of other cool things will appear =) The equation sets direct, parallel to the x-axis and an additional condition (cm. linear inequalities)
tells us that we need to calculate the area of the following figure: 
I will associatively call the desired shaded figure the “roof of the house”, the rectangle – the “wall of the house”, and the entire structure (wall + roof) – the “facade of the house”. Although this building looks more like some kind of cowshed =)
To find the area of the “roof” it is necessary to subtract the area of the “wall” from the area of the “facade”.
First, let's deal with the "facade". To find its area, you need to find out the values that specify the points of intersection of the line with the first arc of the cycloid (points and ). Let's substitute into the parametric equation:
A trigonometric equation can be easily solved by simply looking at cosine plot: on the interval, equality is satisfied by two roots: . In principle, everything is clear, but, nevertheless, let’s play it safe and substitute them into the equation:
– this is the “X” coordinate of the point;
– and this is the “X” coordinate of the point.
Thus, we are convinced that the parameter value corresponds to the point , and the value corresponds to the point .
Let's calculate the area of the "facade". For more compact notation, the function is often differentiated directly below the integral: 
The area of the “wall” can be calculated using the “school” method by multiplying the lengths of adjacent sides of the rectangle. The length is obvious, all that remains is to find it. It is calculated as the difference between the “X” coordinates of the points “tse” and “be” (found earlier):
Wall area:
Of course, there is no shame in finding it even with the help of the simplest definite integral from the function on the segment:
As a result, the roof area is:
Answer: 
And, of course, if we have a drawing, we estimate, box by box, whether the result obtained is similar to the truth. Similar
The following task is for you to solve on your own:
Example 5
Calculate the area of a figure bounded by lines given by equations 
Let us briefly systematize the solution algorithm:
– In most cases, you will have to make a drawing and determine the figure whose area you want to find.
– At the second step, you should understand how the required area is calculated: it can be a single curved trapezoid, it can be a difference in areas, it can be a sum of areas - in short, all those chips that we looked at in the lesson.
– At the third step, we need to analyze whether it is advisable to use the symmetry of the figure (if it is symmetrical), and then find out the limits of integration (the initial and final value of the parameter). Usually this requires solving the simplest trigonometric equation– here you can use the analytical method, the graphical method or the simple selection of the necessary roots according to trigonometric table.
! Don't forget that parametric equations can “draw” a line from right to left, in this case we make an appropriate reservation and amendment in the working formula.
– And at the final stage, technical calculations are carried out. It is always nice to evaluate the plausibility of the answer obtained from the drawing.
And now the long-awaited meeting with the star:
Example 6
Calculate the area of a figure bounded by lines given by equations
Solution: the curve given by the equations is astroid, And linear inequality uniquely identifies the shaded figure in the drawing: 
Let's find the parameter values that determine the intersection points of the line and the astroid. To do this, let’s substitute into the parametric equation:
Methods for solving such an equation have already been listed above; in particular, these roots can be easily selected according to trigonometric table.
The figure is symmetrical about the x-axis, so let's calculate the upper half of the area (blue shading), and double the result.
Let's substitute the value into the parametric equation: 
As a result, we obtained the “Greek” coordinate of the upper (we need) point of intersection of the astroid and the straight line.
The right vertex of the astroid obviously corresponds to the value . Let's check just in case: 
, which was what needed to be checked.
As with the ellipse, the parametric equations “draw” the arc of the astroid from right to left. For variety, I’ll format the ending in the second way: when the parameter changes within the limits, the function decreases, therefore (don’t forget to double!!):
The integral turned out to be quite cumbersome, and in order to “not carry everything around with you,” it is better to interrupt the solution and transform the integrand separately. Standard lower the degree by using trigonometric formulas:
Suitable, in the last term let's put the function under the differential sign:
Answer:
Yes, it’s a little hard with the stars =)
The following assignment is for advanced students:
Example 7
Calculate the area of a figure bounded by lines given by equations
To solve it, there will be enough materials that we have already considered, but the usual path is very long, and now I will tell you about one more effective method. The idea is actually familiar from the lesson Calculating area using a definite integral– this is integration over the “y” variable and the use of the formula 
. Substituting parametric functions into it, we obtain a mirror working formula:
Indeed, why is it worse than the “standard” one? This is another advantage of the parametric form - the equation 
capable of playing the role not only of an “ordinary” one, but simultaneously And inverse function.
IN in this case it is assumed that the functions continuous on the integration interval and the function monotonous On him. Moreover, if decreases on the integration interval (parametric equations “draw” the graph “in the opposite direction” (attention!!) axis), then using the technology already discussed, you should rearrange the limits of integration or initially put a “minus” in front of the integral.
The solution and answer to Example No. 7 are at the end of the lesson.
The final mini-section is devoted to a rarer problem:
How to find the volume of a body of rotation,
if the figure is limited by a parametrically defined line?
Let's update the formula derived at the beginning of the lesson: 
. The general solution method is exactly the same as for finding the area. I’ll pull out a few tasks from my piggy bank.
When we figured out the geometric meaning of a definite integral, we came up with a formula that can be used to find the area of a curvilinear trapezoid bounded by the x-axis and straight lines x = a, x = b, as well as a continuous (non-negative or non-positive) function y = f(x). Sometimes it is more convenient to specify the function that limits the figure in parametric form, i.e. express the functional dependence through the parameter t. In this material, we will show how you can find the area of a figure if it is limited by a parametrically defined curve.
After explaining the theory and deriving the formula, we will look at several typical examples to find the area of such figures.
Basic formula for calculation
Let us assume that we have a curvilinear trapezoid, the boundaries of which are the straight lines x = a, x = b, the O x axis and a parametrically defined curve x = φ (t) y = ψ (t), and the functions x = φ (t) and y = ψ (t) are continuous on the interval α; β, α< β , x = φ (t) будет непрерывно возрастать на нем и φ (α) = a , φ (β) = b .
Definition 1
To calculate the area of a trapezoid under such conditions, you need to use the formula S (G) = ∫ α β ψ (t) · φ " (t) d t.
We derived it from the formula for the area of a curvilinear trapezoid S (G) = ∫ a b f (x) d x by substitution x = φ (t) y = ψ (t):
S (G) = ∫ a b f (x) d x = ∫ α β ψ (t) d (φ (t)) = ∫ α β ψ (t) φ " (t) d t
Definition 2
Taking into account the monotonic decrease of the function x = φ (t) on the interval β; α, β< α , нужная формула принимает вид S (G) = - ∫ β α ψ (t) · φ " (t) d t .
If the function x = φ (t) is not one of the basic elementary ones, then we will need to remember the basic rules for increasing and decreasing a function on an interval to determine whether it will be increasing or decreasing.
In this paragraph we will analyze several problems using the formula derived above.
Example 1
Condition: find the area of the figure formed by the line given by equations of the form x = 2 cos t y = 3 sin t.
Solution
We have a parametrically defined line. Graphically it can be displayed as an ellipse with two semi-axes 2 and 3. See illustration:
Let's try to find the area 1 4 of the resulting figure, which occupies the first quadrant. The region is in the interval x ∈ a; b = 0 ; 2. Next, multiply the resulting value by 4 and find the area of the whole figure.
Here is the progress of our calculations:
x = φ (t) = 2 cos t y = ψ (t) = 3 sin t φ α = a ⇔ 2 cos α = 0 ⇔ α = π 2 + πk , k ∈ Z , φ β = b ⇔ 2 cos β = 2 ⇔ β = 2 πk , k ∈ Z
With k equal to 0, we get the interval β; α = 0 ; π 2. The function x = φ (t) = 2 cos t will decrease monotonically on it (for more details, see the article on the main elementary functions and their properties). This means that you can apply the formula for calculating the area and find the definite integral using the Newton-Leibniz formula:
- ∫ 0 π 2 3 sin t · 2 cos t " d t = 6 ∫ 0 π 2 sin 2 t d t = 3 ∫ 0 π 2 (1 - cos (2 t) d t = = 3 · t - sin (2 t) 2 0 π 2 = 3 π 2 - sin 2 π 2 2 - 0 - sin 2 0 2 = 3 π 2
This means that the area of the figure given by the original curve will be equal to S (G) = 4 · 3 π 2 = 6 π.
Answer: S(G) = 6π
Let us clarify that when solving the problem above, it was possible to take not only a quarter of the ellipse, but also its half - the upper or lower one. One half will be located on the interval x ∈ a; b = - 2 ; 2. In this case we would have:
φ (α) = a ⇔ 2 cos α = - 2 ⇔ α = π + π k, k ∈ Z, φ (β) = b ⇔ 2 cos β = 2 ⇔ β = 2 π k, k ∈ Z
Thus, with k equal to 0, we get β; α = 0 ; π. The function x = φ (t) = 2 cos t will decrease monotonically on this interval.
After this, we calculate the area of half the ellipse:
- ∫ 0 π 3 sin t · 2 cos t " d t = 6 ∫ 0 π sin 2 t d t = 3 ∫ 0 π (1 - cos (2 t) d t = = 3 · t - sin (2 t) 2 0 π = 3 π - sin 2 π 2 - 0 - sin 2 0 2 = 3 π
It is important to note that you can only take the top or bottom, but not the right or left.
You can create a parametric equation for a given ellipse, the center of which will be located at the origin. It will look like x = a · cos t y = b · sin t . Proceeding in the same way as in the example above, we obtain a formula for calculating the area of the ellipse S e l and p with a = πab.
You can define a circle whose center is located at the origin using the equation x = R · cos t y = R · sin t , where t is a parameter and R is the radius of this circle. If we immediately use the formula for the area of an ellipse, then we will get a formula with which we can calculate the area of a circle with radius R: S k r y r a = πR 2 .
Let's look at one more problem.
Example 2
Condition: find what the area of the figure will be equal to, which is limited by a parametrically defined curve x = 3 cos 3 t y = 2 sin 3 t.
Solution
Let us immediately clarify that this curve has the shape of an elongated astroid. Typically the astroid is expressed using an equation of the form x = a · cos 3 t y = a · sin 3 t .
Now let's look in detail at how to construct such a curve. Let's build based on individual points. This is the most common method and is applicable for most tasks. More complex examples require differential calculus to identify a parametrically defined function.
We have x = φ (t) = 3 cos 3 t, y = ψ (t) = 2 sin 3 t.
These functions are defined for all real values of t. For sin and cos it is known that they are periodic and their period is 2 pi. Having calculated the values of the functions x = φ (t) = 3 cos 3 t, y = ψ (t) = 2 sin 3 t for some t = t 0 ∈ 0; 2 π π 8 , π 4 , 3 π 8 , π 2 , . . . , 15 π 8, we get points x 0; y 0 = (φ (t 0) ; ψ (t 0)) .
Let's make a table of the total values:
| t 0 | 0 | π 8 | π 4 | 3 π 8 | π 2 | 5 π 8 | 3 π 4 | 7 π 8 | π |
| x 0 = φ (t 0) | 3 | 2 . 36 | 1 . 06 | 0 . 16 | 0 | - 0 . 16 | - 1 . 06 | - 2 . 36 | - 3 |
| y 0 = ψ (t 0) | 0 | 0 . 11 | 0 . 70 | 1 . 57 | 2 | 1 . 57 | 0 . 70 | 0 . 11 | 0 |
| t 0 | 9 π 8 | 5 π 4 | 11 π 8 | 3 π 2 | 13 π 8 | 7 π 4 | 15 π 8 | 2π |
| x 0 = φ (t 0) | - 2 . 36 | - 1 . 06 | - 0 . 16 | 0 | 0 . 16 | 1 . 06 | 2 . 36 | 3 |
| y 0 = ψ (t 0) | - 0 . 11 | - 0 . 70 | - 1 . 57 | - 2 | - 1 . 57 | - 0 . 70 | - 0 . 11 | 0 |
After this, mark the required points on the plane and connect them with one line.
Now we need to find the area of that part of the figure that is located in the first coordinate quarter. For it x ∈ a; b = 0 ; 3:
φ (α) = a ⇔ 3 cos 3 t = 0 ⇔ α = π 2 + πk , k ∈ Z , φ (β) = b ⇔ 3 cos 3 t = 3 ⇔ β = 2 πk , k ∈ Z
If k is equal to 0, then we get the interval β; α = 0 ; π 2 , and the function x = φ (t) = 3 cos 3 t will decrease monotonically on it. Now we take the area formula and calculate:
- ∫ 0 π 2 2 sin 3 t · 3 cos 3 t " d t = 18 ∫ 0 π 2 sin 4 t · cos 2 t d t = = 18 ∫ 0 π 2 sin 4 t · (1 - sin 2 t) d t = 18 ∫ 0 π 2 sin 4 t d t - ∫ 0 π 2 sin 6 t d t
We have obtained definite integrals that can be calculated using the Newton-Leibniz formula. Antiderivatives for this formula can be found using the recurrent formula J n (x) = - cos x · sin n - 1 (x) n + n - 1 n J n - 2 (x) , where J n (x) = ∫ sin n x d x .
∫ sin 4 t d t = - cos t · sin 3 t 4 + 3 4 ∫ sin 2 t d t = = - cos t · sin 3 t 4 + 3 4 - cos t · sin t 2 + 1 2 ∫ sin 0 t d t = = - cos t sin 3 t 4 - 3 cos t sin t 8 + 3 8 t + C ⇒ ∫ 0 π 2 sin 4 t d t = - cos t sin 3 t 4 - 3 cos t sin t 8 + 3 8 t 0 π 2 = 3 π 16 ∫ sin 6 t d t = - cos t · sin 5 t 6 + 5 6 ∫ sin 4 t d t ⇒ ∫ 0 π 2 sin 6 t d t = - cos t · sin 5 t 6 0 π 2 + 5 6 ∫ 0 π 2 sin 4 t d t = 5 6 3 π 16 = 15 π 96
We calculated the area of a quarter of a figure. It is equal to 18 ∫ 0 π 2 sin 4 t d t - ∫ 0 π 2 sin 6 t d t = 18 3 π 16 - 15 π 96 = 9 π 16.
If we multiply this value by 4, we get the area of the entire figure - 9 π 4.
In exactly the same way, we can prove that the area of the astroid, given by the equations x = a · cos 3 t y = a · sin 3 t, can be found by the formula S a stroid = 3 πa 2 8, and the area of the figure , which is limited by the line x = a · cos 3 t y = b · sin 3 t , is calculated using the formula S = 3 πab 8 .
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