Volume of a solid

[lamaclass]

I was working on this problem and got stuck after a certain point:

Find the volume of the solid generated by revolving the region bounded by the curves y = -x[sup:2crvj3cz]2[/sup:2crvj3cz]+6 and y=3x+2 about the line y=8. Set up only.

Am I on the right track? If I use the shell method:

-x[sup:2crvj3cz]2[/sup:2crvj3cz]+6=3x+2

=(x-1) (x+4)

x= -4 or 1

p(t) = 8-x, h(t) = x[sup:2crvj3cz]2[/sup:2crvj3cz]+3x-4 ?

 

[mmm4444bot]

Since the axis of rotation is horizontal, the shell method will require integrating with respect to y.

If you want to integrate with respect to x (easier, since the curves are given in terms of x), use the disk method (aka washer method).

Do the instructions specifically call for the shell method?

If so, then you don't want x to go from -4 to 1. You want y to go from -10 to 5.

Start by solving each of the two curve equations for x in terms of y.

Otherwise, let us know that you're going to use the disk method, instead.

 

[galactus]

A point to remember when using shells is that the cross sections are parallel to the axis we are revolving about.

Washers is easier.

Using washers, \(\displaystyle {\pi}\int_{-4}^{1}\left[(-x^{2}-2)^{2}-(3x-6)^{2}\right]dx\)

 

[lamaclass]

Ahh I see that now. It said we could solve it either way. So just to clarify, you're going solve with respect to x whenever it's vertical and solve with respect to y whenever it's horizontal right?

I'm going to solve it using the Disc Method in that case.

 

[lamaclass]

Using the Disc Method to it, I got an answer of 1300. Is that correct?

 

[galactus]

Your solution should involve Pi. \(\displaystyle \frac{1000{\pi}}{3}\)

 

[lamaclass]

Oh whoop forgot that in the end! Otherwise for you both for your help!

 

[BigGlenntheHeavy]

\(\displaystyle Washer:\)

\(\displaystyle \pi\int_{-4}^{1}[(3x-6)^{2}-(-x^{2}-2)^{2}]dx \ = \ \frac{1000\pi}{3}.\)

\(\displaystyle Shell:\)

\(\displaystyle 2\pi\int_{5}^{6}2\sqrt{6-y}(8-y)dy+2\pi\int_{2}^{5}\bigg(\frac{y-2}{3}\bigg)(8-y)dy+2\pi\int_{2}^{5}\sqrt{6-y}(8-y)dy\)

\(\displaystyle +2\pi\int_{-10}^{2}\bigg[\bigg(\frac{y-2}{3}\bigg)+\sqrt{6-y}\bigg](8-y)dy \ = \ \frac{1000\pi}{3}.\)

\(\displaystyle Washer \ is \ easier.\)