# 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.