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.