Question

Прямоугольный треугольник с гипотенузой 15 и площадью 36 вращается вокруг большего катета. найти объём и площаль поверхности фигуры вращенич

Answer 1

Объем конуса с радиусом 9 и высотой 12

Answer 2

Problem Analysis

We are given a right triangle with a hypotenuse of 15 and an area of 36. The triangle is rotated around the longer leg. We need to find the volume and surface area of the resulting solid.

Solution

To solve this problem, we can use the formulas for the volume and surface area of a solid of revolution.

Volume of the Solid of Revolution

The volume of the solid of revolution can be calculated using the formula:

V = π * ∫[a,b] (f(x))^2 dx

where: - V is the volume of the solid of revolution, - π is a mathematical constant approximately equal to 3.14159, - ∫[a,b] represents the definite integral over the interval [a,b], - f(x) is the function that represents the shape of the solid when rotated around the axis of rotation, and - dx represents an infinitesimally small change in x.

In this case, the function f(x) represents the length of the cross-section of the solid at a given x-coordinate. Since we are rotating a right triangle around the longer leg, the length of the cross-section will be equal to the length of the longer leg at that x-coordinate.

Let's denote the longer leg of the right triangle as x and the shorter leg as y. We can use the Pythagorean theorem to relate x and y:

x^2 + y^2 = 15^2

We are also given that the area of the right triangle is 36, so we have:

(1/2) * x * y = 36

Solving these two equations simultaneously will give us the values of x and y.

Surface Area of the Solid of Revolution

The surface area of the solid of revolution can be calculated using the formula:

A = 2π * ∫[a,b] f(x) * √(1 + (f'(x))^2) dx

where: - A is the surface area of the solid of revolution, - π is a mathematical constant approximately equal to 3.14159, - ∫[a,b] represents the definite integral over the interval [a,b], - f(x) is the function that represents the shape of the solid when rotated around the axis of rotation, - f'(x) is the derivative of f(x) with respect to x, and - dx represents an infinitesimally small change in x.

In this case, the function f(x) represents the length of the cross-section of the solid at a given x-coordinate, which is equal to the length of the longer leg of the right triangle.

To find the derivative of f(x), we can differentiate the equation x^2 + y^2 = 15^2 with respect to x:

2x + 2y * dy/dx = 0

Simplifying this equation, we get:

dy/dx = -x/y

Substituting the value of y from the equation (1/2) * x * y = 36, we can find the derivative dy/dx in terms of x:

dy/dx = -x / (2 * 36 / x) = -x^2 / 72

Now, we can substitute the values of f(x) and f'(x) into the surface area formula and integrate to find the surface area of the solid of revolution.

Calculation

Let's solve the equations to find the values of x and y:

(1/2) * x * y = 36

x^2 + y^2 = 15^2

From the first equation, we can express y in terms of x:

y = (72 / x)

Substituting this value of y into the second equation, we get:

x^2 + (72 / x)^2 = 15^2

Simplifying this equation, we have:

x^4 + 72^2 = 15^2 * x^2

Rearranging the terms, we get:

x^4 - 15^2 * x^2 + 72^2 = 0

Solving this quadratic equation will give us the values of x. We can then substitute these values into the equation y = (72 / x) to find the corresponding values of y.

Once we have the values of x and y, we can calculate the volume and surface area of the solid of revolution using the formulas mentioned earlier.

Conclusion

To find the volume and surface area of the solid of revolution formed by rotating a right triangle with a hypotenuse of 15 and an area of 36 around the longer leg, we need to solve the equations x^2 + y^2 = 15^2 and (1/2) * x * y = 36 to find the values of x and y. Then, we can use these values to calculate the volume and surface area using the formulas mentioned earlier.

Please note that the exact values of x and y, as well as the volume and surface area, will depend on the solutions of the equations and the accuracy of the calculations.