Finding volume using shell method with region $y=0$, $y=\cos(5x), x=\pi/10, x=0$, rotated about $y = -2$

Question

I'm thinking

(a) the radius = $y+2$

(b) the thickness = $dx$

(c) the height = $x-\frac{\pi}{10}$

(d) the range = $y$: $0\rightarrow1$

(e) $x=\frac{arccos(y)}{5}$

giving $V=2\pi\int_0^1(\frac{\pi}{10}-\frac{arccos(y)}{5})(y+2)dy$

did I set this up wrong? also, I'm not sure the best way to go about integrating this if it's right. would I use substitution?

Answer 1

The integral seems to be correct at all, but

(b) the thickness = dy

(c) the height = $x$

(d) the range = y: $0\rightarrow1$

(e) x= $\frac{\pi}{10}-\frac{arccos(y)}{5}$

For the integration you can use the direct formula for $\arccos y$ and itegrate by parts the term with $y\arccos y$.