Revolving the region $y=6-x^2$, $y=2x+3$ about x=5

Question

To find the volume of the resulting solid we can either use rings or shells (parallel or perpendicular slices) For this particular example using shells (parallel slices) makes it simple: We have $\int (5-x)(6-x^2-2x-3) dx $ going from $x=-1$ to $x=3$ and the rest is simple. But when I attempt to setup the integrals of the inner and outer radii for the ring method : 1)the interval has to be divided to have one integral from $y=-3$ to $y=5$ and another one from $y=5$ to $y=6$ however I fail to setup the inner and outer radii and get the same result, I did a lot of tries involving random math to see if I would get the same result but no avail... TL;DR Can you find the volume using rings, explaining anything that has to be?

Answer 1

Firs of all, the bounds of integration for the shell method should be from $−3$ to $1$ because those are the roots of the equation $6-x^2=2x+3$:

$$ V=2\pi\int_{-3}^{1}\left[(5-x)(6-x^2-(2x+3))\right]\,dx=128\pi\ cubic\ units. $$

Wolfram Alpha check.

And this is your integral for the washer method (I assume that's what you mean by "rings"):

$$ V= \pi\int_{-3}^{5}\left[\left(5-\left(-\sqrt{6-y}\right)\right)^2-\left(5-\frac{y-3}{2}\right)^2\right]\,dy+\\ \pi\int_{5}^{6}\left[\left(5-\left(-\sqrt{6-y}\right)\right)^2-\left(5-\sqrt{6-y}\right)^2\right]\,dy=\\ \frac{344\pi}{3}+\frac{40\pi}{3}= 128\pi\ cubic\ units. $$

As you can see, the integration should be done with respect to $y$. The first integral on Wolfram Alpha and the second one. That's just a check. If there is something you don't understand, don't hesitate to ask. But the main idea with the waster method is that you subtract the smaller volume from the bigger one.

And you do a similar thing for the other piece.

Answer 2

You are right we have to use parallel slices but the integrand should be $2\pi(5 - x)dxdy$. You are also correct to split the volume into two integrals with different limits. The integral should be

$$\int_{5}^{6}\int_{- \sqrt{6 - y}}^{\sqrt{6 - y}}2\pi (5 - x)dxdy + \int_{-3}^{5} \int_{-\sqrt{6 - y}}^{\frac{y - 3}{2}} 2\pi (5 - x)dxdy = 128\pi$$