# Find the area of the region enclosed by $y=\sqrt{x}$, $y=x$, rotated about $x=6$

Call the equations: $1) y=\sqrt{x}$ and $2) y=x$.

I have no idea why I am having such trouble with this question. My guess is that it lies in the fact that I must have the radii wrong using the washer method.

So here is what I have done so far. First, drew a picture and decided that the washer method was the proper technique. Because the curves are being rotated about the line $x=6$, I will need to integrate with respect to $y$ with limits $0$ and $1$ since $\sqrt{x}=x$ when $x=0,1$.

Because I am integrating with respecting to $y$ and I need to re-write my functions $1) y^2 = x$ and $2) y = x$.

Here are my thoughts: I must integrate $A(1) - A(2)$ since $\sqrt{x}$ is the "outside" function. To do this I need to find the radii of equations $1$ and $2$ and plug them into,

$\int_0^1 (A(1)-A(2)) dy$ where $A(i) = \pi r^2$ for $i=1,2$

Because we are rotating around $x=6$ shouldn't the radii just be $6 +$ the function? I know what the answer should be $\frac{28}{15}\pi$, but nothing I have tried has given that answer back to me. Can some please help me understand how to determine the radii here?

Thank you.

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Rotating about $x=6$, the radii would be $6-y$ and $6-y^2$. – David Mitra Mar 4 '12 at 18:45
You mean volume, and not area? – Aryabhata Mar 4 '12 at 18:46
Yes, I meant volume. And thank you David! – Leslie Mar 4 '12 at 18:58

Since you are rotating about the line $x = 6$, when you look back at your drawing you will see that the outer radius is the distance from the line $x=6$ to the curve $x = y^2$. To calculate this you can just use the distance formula to get $r_{out}= 6 - y^2$, and similarly for the inner radius to obtain $r_{in} = 6 - y$. Now your calculation should be
$$\int_0^1 \pi(r_{out})^2 - \pi(r_{in})^2 dy$$
Treble, I don't see how $y=x$ gives you the outer radius. And in fact, the integration only gets you a positive volume if $y=\sqrt{x}$ or $x=y^2$ is the outer radius. – Leslie Mar 4 '12 at 19:00