Find the volume of the solid obtained when the region bounded by $y = \sqrt x, y = −x$ and the line $x = 4$ is revolved around the $x$-axis. Find the volume of the solid obtained when the region bounded by $y = \sqrt x, y = −x$ and the line $x = 4$ is revolved around the $x$-axis.
Here's the figure :
I'm considering ​the volume of the solid as a composition of infinitesimal discs of radius $f(x)=\sqrt x$ from $x=1$ to $x=4$. Note that each disc will be of volume $\pi (dr)^2 dh$ where $r=\sqrt x$ at the particular value of $x$.
From $0$ to $1$ the volume will be the result of rotating the graph under $\sqrt x$ and then from $1$ to $4$, we will have the volume obtained from rotating $y=x$. So, how do we canculate the volume from $0$ to $1$? $\int_{1}^{4} \int_{1}^{4} \pi (\sqrt x)^2 dx dh $. I think this is correct. Isn't it?
 A: You have to split this up into two regions, I think, based on which function is "farther out":
$$V=\pi\int_0^1 x \, dx + \pi\int_1^4 x^2\,dx.$$
This is an unusual problem, because of the axis of rotation going through the region. Most textbook problems have the axis of rotation on the boundary, or outside the region entirely.
A: This is a beautiful problem.. Observe that rotating the line $y=-x$ in the $3$ dimensions in any direction, keeping the angle made by it with the $x$ axis constant $(\frac{π}4)$ won't affect our volume.. Just for the sake of seeing the situation clearly, let's consider $y=x$ in place of $y=-x$.
Note that you could have considered $y=-x$ too, there's absolutely no problem, but this one will be easier for me to explain :)
Now since everything comes in the very first quadrant, observe what you have, the graph of $y=\sqrt{x}$ and $y=x$. Sorry, I do not know how to attach graphs, so consider it in your head (it's easy anyway) and bear with me :P. These two graphs intersect at $x=1$. When $x\in(0,1)$ the curve of $y=\sqrt{x}$ lies above $y=x$ and therefore engulfs it's little brother when both of these are rotated with respect to the $x$ axis to constitute the volume..

This is exactly why I loved this problem at once when I saw it!

So, it's clear to you that when $x\in(0,1)$ the graph of $y=\sqrt{x}$ has a greater say in the volume than $y=x$, while just the opposite happens when $x\in(1,4)$. There the graph of $y=x$ is above $y=\sqrt{x}$. So, it reduces to a very simple problem of calculating the two volumes separately, and due to continuity at the point $1$ adding them up is completely legal..
So, coming to calculations you may use any method to calculate volumes.. I here use the disk method, by which I conclude the volume must be -
$$π\int_{0}^{1}xdx+π\int_{1}^{4}x^2dx$$ Which you may solve easily.. The answer should be $\frac{43π}2$.
