TITLE: Find volume of rev using washer AND shell
QUESTION [0 upvotes]: Volume of revolution for the region bounded by $y= x^2$, $y=0$, $x=1$, $x=2$ about the line $x=4$.
The shell method is straightforward and I get the correct answer of $67 \pi /6$.
However, working this via disk-washer appears to need 2 separate regions - one from $0$ to $1$, and the second from $1$ to $4$. I don’t get the same answer, instead getting $11 \pi /3$. 
Can someone assist with the disk-washer method for this problem?

REPLY [0 votes]: Yes, you have to set up  two integrals.
The bottom part  is $$\int _0^1 \pi (9-4) dy$$
and the top part is $$ \int _1^4 \pi ((4-\sqrt y)^2 -4) dy$$

REPLY [0 votes]: The washer integral is 
$$\pi\int_0^1 (3^2-2^2)dy 
+ \pi \int_1^4 [(4-\sqrt y)^2- (4-2)^2]dy$$
$$=5\pi +\frac{37\pi}6=\frac{67\pi}6$$
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