# Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 20

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Use the shell method to find the volume generated by rotating the region bounded by the curves $y = x^2, x = y^2$ about the $y = -1$. Sketch the region and a typical shell.

By using a horizontal strip, notice that the distance of the strips from the line $y = -1$ is $1 + y$. If you revolve this distance about $y = -1$, you'll have a circumference of $C = 2 \pi ( 1 + y)$. Also, notice that the height of the strips resembles the height of the cylinder as $H = x_{\text{right}} - x_{\text{left}} = \sqrt{y} - y^2$. Thus, we have

$\displaystyle V = \int^b_a C(y) H(y) dy$

\begin{aligned} V =& \int^1_0 2 \pi (1 + y) (\sqrt{y} - y^2) dy \\ \\ V =& 2 \pi \int^1_0 (\sqrt{y} - y^2 + y^{\frac{3}{2}} - y^3 ) dy \\ \\ V =& 2 \pi \left[ \displaystyle \frac{y^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} - \frac{y^3}{3} + \frac{y^{\frac{5}{2}}}{\displaystyle \frac{5}{2}} - \frac{y^4}{4} \right]^1_0 \\ \\ V =& \frac{29 \pi}{30} \text{ cubic units} \end{aligned}