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I have 2 lines, $y=x^2$ and $x=y^2$, and I am trying to solve for the volume of the solid created by rotating the region bounded by those 2 lines around the line $x=-1$.

The region bounded by these 2 lines looks somewhat like this (closest image I could find):

enter image description here

This region is then rotated around the line $x=-1$ which creates a bowl-like solid.

I know how to solve for the volume using $\int^b_a f(x)-g(x)\,\mathrm{d}x$, but I am getting tripped up by the fact that these two lines are functions of x and y. This is what I tried:

$$\begin{align*} V &= \int^1_0 \pi(\sqrt y)^2-\pi(y^2)^2\,\mathrm{d}y\\ V &= \pi\int_0^1 y - y^4\,\mathrm{d}y\\ V &= \pi[\frac{1}{2}y^2-\frac{1}{5}y^5]|_0^1\\ V &= \frac{1}{2}\pi - \frac{1}{5}\pi\\ V &= \frac{3}{10}\pi \end{align*}$$

Which is incorrect. What am I doing wrong?

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The inner and outer radii aren’t $\sqrt{y}$ and $y^2$, because your axis of revolution isn’t the $y$-axis. The axis of revolution is at $x=-1$, one unit further away than your calculation makes it. HINT: If you were using shells instead of washers, the radius of the shell at $x$ would be $x-(-1)=x+1$, not $x$. The method of washers also requires an adjustment, though the details are different. –  Brian M. Scott Feb 9 '12 at 5:08
    
So I have to add one to each of those?: $\sqrt y + 1$ and $y^2 + 1$? –  Marlon Feb 9 '12 at 5:10
    
Yes, before you square them. –  Brian M. Scott Feb 9 '12 at 5:11
    
@BrianM.Scott Wow thank you! I got the correct answer! If you make your comment an answer I will select it. –  Marlon Feb 9 '12 at 5:18

1 Answer 1

up vote 3 down vote accepted

The inner and outer radii aren’t $\sqrt{y}$ and $y^2$, because your axis of revolution isn’t the $y$-axis. The axis of revolution is at $x=-1$, one unit further away than your calculation makes it.

HINT: If you were using shells instead of washers, the radius of the shell at $x$ would be $x-(-1)=x+1$, not $x$. The method of washers also requires an adjustment, though the details are different.

Always remember that you’re interested in the distance to the axis of revolution, wherever it is, and not necessarily to the coordinate axis, and you should be okay.

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Thanks again. (That last part you added cleared up another question of mine too)! –  Marlon Feb 9 '12 at 5:27

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