Shell method for volume

[kheath39]

Use the shell method to find the volume of the solid generated by revolving the region bounded by the line y=x+2 and the parabola y=x^2 about the following line. y=4


I can get the correct answer by doing the washer method, but no matter how I manipulate this I cannot get the answer using the shell method.

Ive set it up as upper b 4 lower b 1 2pi(4-y)(sqrty-(y-2)dy + upper b 1 lower b 0 2piy(2sqrty)dy and that is incorrect. I've tried so many other things but cannot get it. I've used my online tutoring service and have burned through 8 "tutors" that cannot get it.

The correct answer is 108pi/5 and can be solved via the washer method upper b 2 lower b -1 pi((4-x^2)^2-(4-(x+2))^2dx

 

[stapel]

Use the shell method to find the volume of the solid generated by revolving the region bounded by the line y=x+2 and the parabola y=x^2 about the following line. y=4

Ive set it up as upper b 4 lower b 1 2pi(4-y)(sqrty-(y-2)dy + upper b 1 lower b 0 2piy(2sqrty)dy and that is incorrect....

How did you arrive at these boundaries and integrands?

Since this is being rotated around y = 4 and you have to do by shells (that is, by hollow cylinders, which are circles multiplied by heights), you'll be working in terms of dy, rather than in terms of dx. You'll start (after the graphing) by converting things from "y in terms of x" to "x in terms of y".

. . . . .\(\displaystyle y\, =\, x\, +\, 2\, \Rightarrow\, y\, -\, 2\, =\, x\)

. . . . .\(\displaystyle y\, =\, x^2\, \Rightarrow\, \pm \sqrt{y\, }\, =\, x\)

Doing the graph (and turning it sideways, if this helps), you can see that you'll be using one integrand from y = 0 to y = 1; then, when the straight line comes into play, you'll be using a second integrand from y = 1 to y = 4.

Volumes by shells is done (here) by height of the cylinder (being the top function, less the bottom function), and multiplying by the radius and 2pi. That is, you'll find the circumference of the circle for the cylinder, and multiply by the height.

Using this setup, I get the correct value. Please reply showing all of your steps for this. Thank you!

 

[kheath39]

I got it, thank you.

On my second integral for the radius I was no longer taking 4-y and just changed it to y and overlooked it as I tried a million other things.

Its always the simple things.