Question

Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^3, x = 2, and y = 1 about the y-axis.

93pi/5 <== I put this and got it wrong
120pi/7
47pi/5
None of these

Answer 1

Since the rotation is about the y-axis, I'll integrate by dy.

`\displaystyle y=x^3\\x=\sqrt[3]y\\\\V=\pi \int \limits_1^8(2^2-(\sqrt[3]y)^2)\, dy\\V=\pi \Big[4x-(3)/(5)x^{\tfrac{5}{3}}\Big]_1^8\\V=\pi \left(4\cdot8-(3)/(5)\cdot8^{\tfrac{5}{3}\right-\left(4\cdot1-(3)/(5)\cdot1^{\tfrac{5}{3}\right)\right)\\V=\pi \left(32-(96)/(5)-\left(4-(3)/(5)\right)\right)\\V=\pi \left((64)/(5)-(17)/(5)\right)\\V=(47\pi)/(5)`

Answer 2

`(47)/(5)\pi`

Explanation:

To find the volume of the solid of revolution created by rotating the region bounded by the curve y = x³, x = 2, and y = 1 about the y-axis, we can use the method of cylindrical shells.

As the region of revolution is bounded by the graphs of two functions, y = x³ and y = 1, the volume of a solid of revolution using cylindrical shells is given by the integral:

`\displaystyle V=2\pi \int^b_a x(f(x)-g(x))\; \text{d}x`

where:

• a and b are the lower and upper bounds along the x-axis.
• f(x) is the upper function.
• g(x) is the lower function.

The height of the shell, f(x) - g(x), is determined by the vertical distance between the curve y = x³ and the line y = 1. Therefore, the height is

`f(x) - g(x) = x^3 - 1`

The upper bound of the integral is b = 2, since the region is bounded by x = 2.

The lower bound is the x-value of the point of intersection of the two functions y = x³ and y = 1.

These two functions intersect when x = 1. Therefore, the lower bound of the integral is a = 1.

Substitute these values into the integral:

`\displaystyle V=2\pi \int^2_1 x(x^3-1)\; \text{d}x`

Expand the brackets:

`\displaystyle V=2\pi \int^2_1 (x^4-x)\; \text{d}x`

Evaluate the integral by using the power rule: Increase the power by 1, then divide by the new power.

`\displaystyle V=2\pi\left[(x^5)/(5)-(x^2)/(2)\right]^2_1`

`\displaystyle V=2\pi\left[\left(((2)^5)/(5)-((2)^2)/(2)\right)-\left(((1)^5)/(5)-((1)^2)/(2)\right)\right]`

`\displaystyle V=2\pi\left[\left((32)/(5)-(4)/(2)\right)-\left((1)/(5)-(1)/(2)\right)\right]`

`\displaystyle V=2\pi\left((32)/(5)-(4)/(2)-(1)/(5)+(1)/(2)\right)`

`\displaystyle V=2\pi\left((31)/(5)-(3)/(2)\right)`

`\displaystyle V=2\pi\left((62)/(10)-(15)/(10)\right)`

`\displaystyle V=2\pi\left((47)/(10)\right)`

`\displaystyle V=(47)/(5)\pi`

Therefore, the volume of the solid formed by revolving the region bounded by the graphs of y = x³, x = 2, and y = 1 about the y-axis is:

`\Large\boxed{\boxed{(47)/(5)\pi}}`