# Homework Help: Finding volume by integration

1. Feb 21, 2008

http://img408.imageshack.us/my.php?image=picture20017hu6.jpg

The attachement contains both problem and the attemp at a solution.
Specifically, I am stuck on finding the equation for the hight for each solid.
(height "a" for both solids)
Should I just set up a random letter to indicate the height
or is there any way that I can figure out the equation for it?

#### Attached Files:

• ###### Picture 017.jpg
File size:
18.8 KB
Views:
165
Last edited: Feb 21, 2008
2. Feb 21, 2008

### rocomath

Link it somewhere like imageshack, take like a day to get your attachment approved!

3. Feb 21, 2008

Can you see it now?

4. Feb 21, 2008

### rohanprabhu

take the 'O' as origin. Now, 'scan' through the x-axis.. like.. take a distance 'x' from the origin. From there, assume a cuboid of thickness 'dx' and find the area of that and then integrate within limits. The problem is that to find the width and length of the cuboid, you're gonna need the equation of the curve shown there.. which doesn't seem to be given or since the scan is small.. i can't see it.

5. Feb 21, 2008

I think the curve should be a circle becuase the radius is indicated as "r" in the picture.

6. Feb 21, 2008

### <---

I think he means the plane curve, as you will need to know how the slope rises and falls to know the volumes of your pieces.

7. Feb 22, 2008

### rohanprabhu

oh.. ur right mate.. should've figured that out :d

neways.. now that you have a circle, you can easily determine what the length of the cuboid will be at a distance of 'x' from the origin along the line. Also, the breadth is 'dx' and you can use ur image to get the height of the cuboid for a particular 'x'. [since the angle 'α' remains constant, you have a relation between 'x' and the height of the cuboid].

Find that and integrate within proper limits.. you should get the answer. Post here in case you have any problems.

8. Feb 22, 2008

Okay.. I think I got the second one (with right triangle slice)
The "base" of the triangle should be sqrt((R^2-x^2)) (<- I got this from x^2+y^2=R^2)
and the "height" of the trangle is gonna be sqrt((R^2-x^2))*tanα
(since tanα = height/base.. and you know "base" from above.. so you solve for height)

Than I pluged in two equations into
V=1/2*base*height*dx
and ended up with V= 1/2 (R^2-x^2)tanα dx

The rest of the integrating is not a big deal.. so i'll skip:)

Sorry that my work is kind of messy. I should have scanned my work
but I don't have a scanner with me at the moment..

Last edited: Feb 22, 2008
9. Feb 22, 2008