Volume of a solid using Washer Method

Question

The question is as follows

I'm pretty sure you have to use the washer method because the cross-section of the volume is the outer circle - inner where the inner circle: (y = 4) - (y = 1) = y = 3, but I'm having a problem determing what to use for the outer circle.

I drew this picture:

I know that if you subtract y = 1 by y=x^2 you will get the shaded area. Thus I think the radius for the shaded area is

1 - x^2

I confirm this by plugging in x=0 and x = 1 and find that:

x = 0, r = 1 - 0^2 
       r = 1

x = 1, r = 1-1^2
       r = 0

Which makes sense that the length at the middle is just 1 and at the ends is 0. Thus I think the radius of the outer circle is:

radius of outer = 3 + length of shaded area
                = 3 + (1-x^2)
                = 4 - x^2

Thus the integral looks like:

 pi * integral from [-1,1] of (4-x^2)^2 - 3^2 dx

I integrated this, but apparently it is the wrong answer. Any ideas where I went wrong?

Answer 1

The rotation is about $y=5$. Your notation is hard to follow, but for some reason you seem to be rotating about $y=4$.

With the washer method you want: $\pi \int_{-1}^{1}[(5-x^2)^2-4^2]dx$.

With the shell method you want: $2\pi \int_{0}^{1}(5-y)(2\sqrt{y})dy$

Both evaluate to $\frac{4^2\times 11}{15} \pi$.