Volume of Solid from 0-5 Rotated about y=8

[mshiddensecret]

Homework Statement

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

y=http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmsy10/alpha/144/char70.png x−1[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png y=0[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png x=5;

about the line y = 8.

Homework Equations

The Attempt at a Solution

I tried using the washer disk method and got the integral of:

(pi)(8-sqrt(x-1))^2dx and integrating it from x = 0-5.

Doesn't work. [/B]

 

[mfb]

What do you mean with "doesn't work"? That integral is certainly calculable.

 

[Dick]

sqrt(x-1) isn't even defined at x=0. Draw a sketch of the function and boundaries and rethink this. They really are washers, not disks.

 

[mshiddensecret]

I'm confused. Don't know what you mean.

 

[HallsofIvy]

What, exactly, do you not understand? What is \sqrt{x- 1} when x= 0 or 1/2? What does the graph of y= \sqrt{x-1} look like?

 

[mshiddensecret]

I got it with this:

(pi)(8)^2-(pi)(8-sqrt(x-1))^2dx

 

[HallsofIvy]

You still have completely missed the point! But I am hoping that your integration "from x= 0- 5" was a simple misprint. If x< 1, then x-1 < 0 so \sqrt{x- 1} is not a real number. If y= \sqrt{x- 1} then x= y^2- 1. That is a parabola, opening to the right with vertex at (1, 0). Your integration must be from x= 1 to 5, not 0 to 5.