Find the volume of the solid obtained by rotating the region A in the figure about x=3.

Question

I've been having some issues with this problem

I attempted to find the solution using the Shell Method and ended up with 81pi/2 however I used another method and the result was 99pi/2.

Answer 1

If you use the shell method, the integral is

$ V = \displaystyle 2 \pi \int_{x = 0}^3 (3 - x) (9 - x^2) dx = \dfrac{135}{2} \pi $

If you use the disc/washer method, the integral is

$ V = \displaystyle \pi \int_{y = 4}^{13} (9 - (3 - \sqrt{y - 4} )^2 ) dy = \pi \int_{y=4}^{13} ( 6 \sqrt{y - 4} - y + 4 ) \ dy = \pi \bigg[ 4*27 - 1/2 (9)^2 \bigg] = \dfrac{135}{2} \pi $

Answer 2

This is a volume integral in spherical coordinates. I'll move everything 3 units to the left, so we're integrating around the y-axis, to make the math easier. The function becomes $y = (x+3)^2 + 4$. At $x=0$ with our new function, $y=9$, which will be an upper integration bound on y. On x, we need our inner/lower integration bound in terms of y, so the new equation for the curve must be rearranged such that $x = \sqrt{y - 4} -3$. We can finally write the integral:

$$\int^0_{\sqrt{y - 4} -3}\int^9_4\int^{2\pi}_0 xdxdyd\theta$$

The extra x in the integrand is a characteristic of integrals in cylindrical coordinates (just trust it!). We can easily integrate over $\theta$ to get a multiplier of $2\pi$, giving:

$$2\pi\int^0_{\sqrt{y - 4} -3}\int^9_4 xdxdy$$

Since the integral for x has y in it, we have to integrate x first:

$$2\pi\int^9_4 \left.\left[\frac{x^2}{2}\right]\right|^0_{\sqrt{y - 4} -3}dy$$

Evaluating the integral bounds gives:

$$2\pi\int^9_4 \left[0-\frac{\left(\sqrt{y - 4} -3\right)^2}{2}\right]dy$$

... and simplifying:

$$2\pi\int^9_4 \left[-\frac{1}{2}\left(y - 4 - 6\sqrt{y - 4} + 9\right)\right]dy$$

$$2\pi\int^9_4 \left[-\frac{1}{2}\left(y - 6\sqrt{y - 4} + 5\right)\right]dy$$

Taking the $-\frac{1}{2} out front$:

$$-\pi\int^9_4 \left(y - 6\sqrt{y - 4} + 5\right)dy$$

Finally, the last integral can be evaluated:

$$-\pi\left.\left[\frac{y^2}{2} - 6(\frac{2}{3})(y - 4)^{3/2} + 5y\right]\right|^9_4$$

$$-\pi\left[\left(\frac{81}{2} - \frac{16}{2}\right) - \left(4\left((9 - 4)^{3/2}-0^{3/2}\right)\right) + \left(5(9-4)\right)\right]$$

$$-\pi\left[\left(\frac{65}{2} \right) - \left(4(5)^{3/2}\right) + \left(5(5)\right)\right]$$

$$-\pi\left[\frac{65}{2} - 4\left(5\sqrt{5}\right) + 25 \right]$$

$$-\pi\left[\frac{65}{2} - 20\sqrt{5} + 25 \right]$$

We could stop here, or simplify further:

$$-\pi\left[\frac{65}{2} - 20\sqrt{5} + \frac{50}{2} \right]$$

$$-\pi\left[\frac{115}{2} - 20\sqrt{5} \right]$$

$$-\pi\left[\frac{\sqrt{13225}}{2} - 20\sqrt{5} \right]$$

$$-\pi\left[\frac{\sqrt{2645}\sqrt{5}}{2} - 20\sqrt{5} \right]$$

$$-\pi\left[\frac{\sqrt{2645}\sqrt{5}}{2} - \frac{40\sqrt{5}}{2} \right]$$

$$-\pi\left[\frac{\sqrt{2605}\sqrt{5}}{2} \right]$$

$$-\frac{\sqrt{13025}}{2}\pi$$

I assume my negative sign is from my integral bounds, so I'll ignore it and say my result is approximately equal to 179.27. Apologies for any sloppy math errors, but I'm exhausted at this point!