Q:Find the volume of the solid that results when the region enclosed by the given curves is revolved about the $y$-axis:
$$y=\sqrt{\frac{1-x^2}{x^2}}\:(x>0),x=0,y=0,y=2$$
My first problem is I can't imagine the region and the answer provided by the book is: $\pi \tan^{-1}2$ which is far away from what I figured out. Any hints or solution will be appreciated.
Thanks in advance.
Best Answer
Here's the graph of the function (in red) with the area to be revolved around the $y$-axis shaded in blue.
There are two ways you could do this: by re-expressing the function in terms of $y$ and using the disc method, or by leaving the function the way it is and using the shell method.
Using the first method, you would need to evaluate the integral $$\pi\int_0^2[g(y)]^2dy,$$ where $g(y)$ is the same function, but expressed in terms of $y$ (in other words, $x=\ldots$ instead of $y=\ldots$).
To find the volume using the second method, you would need two integrals (can you tell why?): $$2\pi\int_0^ax(2)dx+2\pi\int_a^1xf(x)dx,$$ where $x=a$ is the point of intersection between the function and the line $y=2$, which I'll leave to you to calculate.
Here's a brief explanation of why I used two integrals in the second method:
The general formula for finding the volume of a solid of revolution with the shell method is $$2\pi\int_a^bxf(x)\,dx,$$ where $f(x)$ is some function that gives the height of the region you want to revolve around the $y$-axis. In this case, from $x=0$ to $x=1/\sqrt{5}$, that function is just $y=2$, while from $x=1/\sqrt{5}$ to $x=1$, the height is given by $y=\sqrt{\frac{1-x^2}{x^2}}$. Because the two regions have different heights, we need to use separate integrals to find the volume obtained from revolving each region around the $y$-axis.
In general, any time you see a sharp corner like the one at $(1/\sqrt{5},2)$, that's a sign that you'll need multiple integrals.