Rotating Solids and Integrals

So we are rotating solids. I was given the following information:

$$R=\left\{(x,y) : 1 \le y \le \sqrt{4-x^2}, 1 \le x \le 2\right\}$$

I am supposed to find the volume of the solid when we rotate $R$ around the $y$-axis. I drew the graph and I'm pretty sure the endpoints of the integral are $1$ and $3$. I know I am supposed to use the equation $\pi r^2$ and I'm pretty sure this is a washer. I keep getting a negative area, but I can't figure out what I'm doing wrong.

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The region we are rotating is below the top half of the circle, above the line $y=1$, and to the right of the line $x=1$. The line $x=1$ meets the top half of the circle at $y=\sqrt{3}$.

For the method of washers, you will be integrating with respect to $y$, with $y$ going from $1$ to $\sqrt{3}$.

At height $y$, the washer has inner radius $1$. and outer radius $\sqrt{4-y^2}$.

The volume is $$\int_1^{\sqrt{3}} \pi\left((4-y^2)-1\right)\,dy.$$

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For lower limit, is it 1 instead of zero?:) – juniven Dec 13 '12 at 6:03
@jun: Yes, thank you, I thought it was region above $x$-axis. Luckily, only minor fix needed! – André Nicolas Dec 13 '12 at 6:13
I keep getting a negative number for the area. The indefinite integral of (4-y^2-1) is 3y-1/3*y^3 correct? – Lizi Dec 13 '12 at 6:19
The answer is $\frac{(6\sqrt{3}-8)\pi}{3}$, which is positive. We are talking volume not area.:) – juniven Dec 13 '12 at 6:25
@Lizi: Your indefinite integral is right, so something must have gone wrong in the substitution process. You should get something equivalent to the answer of jun in the preceding comment. – André Nicolas Dec 13 '12 at 6:34

Not knowing how you set it up, I can’t really say where you went astray, but it sounds as if you’re either setting it up wrong or, at best, setting it in what seems to me the slightly harder way to set it up. (The actual mechanics of the calculation may be easier if you set it up to use washers, however.)

Given the way that $R$ is described, I’d use shells, one for each value of $x$ from $1$ to $2$. Since you’re revolving $R$ about the $y$-axis, the radius of the shell at $x$ is just $x$, so the circumference is $2\pi x$. The height of the shell is the distance between its top edge and its bottom edge. The top edge is on the semicircle $y=\sqrt{4-x^2}$, and the bottom edge is on the line $y=1$, so the height of the shell at $x$ is $\sqrt{4-x^2}-1$. Finally, since I’m taking vertical shells, the thickness of the shell at $x$ is $dx$. By the standard formula the volume differential is therefore

$$dV=2\pi x\left(\sqrt{4-x^2}-1\right)dx\;,\tag{1}$$

and the total volume of the solid of revolution is obtained by ‘summing’ (i.e., integrating) $(1)$ over the interval from $x=1$ to $x=2$:

$$V=2\pi\int_1^2x\left(\sqrt{4-x^2}-1\right)dx\;.$$

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Try this.

$V=\int_{1}^{\sqrt{3}}\pi[R^2-r^2]dy$

where $R=\sqrt{4-y^2}$ and $r=1$. Here, we used the Washer Method of obtaining the volume of the solid of revolution.

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My answer is $\frac{(6\sqrt{3}-8)\pi}{3}$. – juniven Dec 13 '12 at 6:23