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

$y=x^2$, $x=y^2$ ; rotated about $y=1$

I know how to sketch the graph. I assumed the integral would be:

$$\pi \int_0^1 [(x^2)^2 - (\sqrt x)^2 ] \, dx $$

I checked the solution and it's actually:

$$\pi \int_0^1 [(1-x^2)^2 - (1-\sqrt x)^2 ] \, dx $$

I know this must have something to do with the rotation about $y=1$ but that's it. Would the integral be: $$\pi\int_0^1 [(1+x^2)^2 - (1+\sqrt x)^2] \, dx$$ if I were to rotate it by $y=-1$? All answers will be appreciated.

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Almost. It would be $$ \pi \int_0^1 [(1+\sqrt x)^2 - (1+x^2)^2 ] \, dx $$ if the axis of rotation were $y = -1$. It's always $$ \pi (\text{outer radius})^2 - \pi (\text{inner radius})^2 $$

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    Basically, whatever the solid is rotated by, add or subtract that from the integral? – 2017-01-30
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    @Dr.Von I probably wouldn't think of it in those terms exactly since it could lead to confusion as to which one gets subtracted from the other. Plus it's technically always subtraction. The $1 + \sqrt x$ and $1 + x^2$ in this answer are really $\sqrt x - (-1)$ and $x^2 - (-1)$. Always do the larger one minus the smaller one. This is why rotating around $y=1$ gives $1 - x^2$ and $1 - \sqrt x$. Because $y=1$ is above or on (i.e., larger than or equal to) both $y=x^2$ and $y=\sqrt x$ for $0 \le x \le 1$. – 2017-01-30