Homework Help: Finding the Volume

1. Oct 27, 2014

mshiddensecret

1. The problem statement, all variables and given/known data

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

about the line y = 8.
2. Relevant equations

3. 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.

Last edited by a moderator: May 7, 2017
2. Oct 27, 2014

Staff: Mentor

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

3. Oct 27, 2014

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.

4. Oct 27, 2014

mshiddensecret

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

5. Oct 27, 2014

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?

6. Oct 27, 2014

mshiddensecret

I got it with this:

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

7. Oct 28, 2014

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.