I need to find the volume of the solid obtained by rotating the region bounded by the following functions:
$xy = 1, y = 0, x = 1, x = 2;$ about $x = -1$
I think the outer radius of the washer is $\pi(2+1)^3$ and the inner radius is $\pi(1+1)^2$ but I'm not sure..I'm really having trouble understanding these type of questions.
Also, I have no idea how to find the upper limit of integration, since I need to integrate with respect to $y$. I think the lower limit is $0$ but I'm not sure about this as well. Can someone please help me find the upper limit and confirm if the outer & radii are correct? Please and thank you.
The outer radius is $2-(-1)=3$ of the washer; $\pi\cdot 3^2=9\pi$ is the area of the disk with that radius. Similarly, $1-(-1)=2$ is the inner radius of the washer, so $\pi\cdot2^2=4\pi$ is the area of the hole in the washer. The actual area of the washer is therefore $=9\pi-4\pi=5\pi$.
However, this is true only when $0\le y\le\frac12$: if you sketch the region (and that should be your first step), you’ll see that when $y>\frac12$, the outer radius of the washer is not $2$, because the outer edge of the region is now on the hyperbola $xy=1$ instead of the line $x=2$. At height $y$, therefore, the outer edge is at $x$-coordinate $x=\frac1y$, and the outer radius of the washer is
$$\frac1y-(-1)=\frac1y+1\;.$$
The inner radius is still $2$, so the area of the washer is now
$$\pi\left(\frac1y+1\right)^2-4\pi=\pi\left(\frac1{y^2}+\frac2y-3\right)\;.$$
Since the thickness of a washer is $dy$, the contribution of the washer at height $y$ to the volume is $dV=5\pi\,dy$ when $0\le y\le\frac12$ and
$$\pi\left(\frac1{y^2}+\frac2y-3\right)dy$$
when $\frac12\le y\le 1$. (The region reaches its maximum height of $1$ at $x=1$.) Thus, you will need two integrals to cover the entire region:
$$V=\int_0^{1/2}5\pi\,dy+\int_{1/2}^1\pi\left(\frac1{y^2}+\frac2y-3\right)dy\;.$$
Note that if you calculate the volume using cylindrical shells, you need only one integral: the height of the shell at $x$-coordinate $x$ is simply $\frac1x$ for all $x$ such that $1\le x\le 2$. Just be a little careful: the radius of that shell is $x-(-1)=x+1$, not $x$.