Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the sold and a typical disk or washer.

y= x^3, y=0, x=1; about the x=2

Answer 1

Since you're rotating around a line parallel to the y axis, you'll want to find the radii at y.
The outer radius will be the cube root of y, and the inner radius will always be one.
The lower bound is 0, the upper bound is 1 (intersection of x = 1 and y = x³)
Plugging into the equation for volume we get:
\[\pi \int\limits_{0}^{1}((2-\sqrt[3]{y})^2 - 1)dy\]

Simplify and integrate.

Answer 2

Okay now how do I simplify this and integrate it properly?

Answer 3

\[\pi\int\limits_{0}^{1}(2-y^\frac{ 1 }{ 3 })^2-1dy\]
\[=\pi\int\limits\limits_{0}^{1}(4- 2y^\frac{ 1 }{ 3 } + y^\frac{ 2 }{ 3 }-1)dy\]
\[=\pi\int\limits\limits\limits_{0}^{1}(3- 2y^\frac{ 1 }{ 3 } + y^\frac{ 2 }{ 3 })dy\]
\[=\pi(3y - \frac{ 3 }{ 4 } y^\frac{ 4 }{ 3 } +\frac{ 3 }{ 5 }y^\frac{ 5 }{ 3 })\]
From 0 to 1
\[=\pi(3 - \frac{ 3 }{ 4 } +\frac{ 3 }{ 5 })\]
\[=2.85\pi\]