# Trying to solve Disk and Washer differently

Here is the question: Find the volume of the region bounded by the $$x$$-axis, $$x=4$$, and $$y=sqrt(x)$$. (rotated about x-axis)

I understand that we can find the cross sectional area at each point with respect to x and integrate ($$\int_{0}^{4}(\pi *x)dx$$), but I want to solve this question slightly differently. Since the curve goes from 0 to 2 (y-axis), I decided to set up my integral like this: $$\int_{0}^{2}(\pi*y^2)dy$$. Unfortunately, this does not give me the correct answer. What went wrong?

Since the rotation axis is the x, then
- in the first (disks) you have to integrate $$\pi y(x)^2 dx$$,
- in the second (shells) you have to integrate $$2 \pi (4-x(y)) dy$$

Have a look at this other post

• I think you are misunderstanding my question. I am referring to the fact that since we know the bounds of the y as being from 0 to 2, why do we have to even consider the equation in terms of x? Why can't we view this as a sum of pi*y^2 where y goes from 0 to 2? – Dude156 Oct 7 '18 at 23:08
• did you have a look to the other similar post I indicated ? there is a sketch that might help to clear the misunderstandings – G Cab Oct 7 '18 at 23:18

Your $$\pi x$$ just so happens to equal $$\pi (\sqrt{x})^2$$ for the disc method. For $$\pi y^2$$, it does not equal $$2\pi y(4 - y^2)$$ for the shell method.

• But why can't we view this problem as the addition of a bunch of disks going from radius 0 to 2? Since this function is one to one, shouldn't this method give us the same answer? – Dude156 Oct 7 '18 at 23:11
• yes,it is a bunch of disks with radius $y$ going from $0$ to $2$: but the disks are piled on $x$, and have thickness of $dx$ – G Cab Oct 7 '18 at 23:24
• @GCab Isn't this an arbitrary chosen quantity? Since both dy and dx are approaching 0, shouldn't we be able to use either one? Also can we continue this conversation in chat.stackexchange.com/rooms/84163/calc-question – Dude156 Oct 7 '18 at 23:27
• If you make a graph and slice your volume so as to make a Riemann sum, you' catch the difference. – G Cab Oct 7 '18 at 23:31
• For the disc method, the volume of a disc is its radius squared times $\pi$ times its thickness which is $\pi r^2 dx(thickness)$ and for the shell method it's the circumference of a shell times its width times its thickness which is $2\pi y(4 - y^2) dy$ – Phil H Oct 7 '18 at 23:38