# Double (volume) integral

1. Oct 5, 2008

### t_n_p

1. The problem statement, all variables and given/known data
- Sketch the region V of 3-space that is bounded above and below by the two surfaces z=Z1(x,y) = 0 and z=Z2(x,y)=1+x^2+y^2 and where the domain of these functions is the region R in the xy plane enclosed by the four lines y=x, y=-x, y=2+x and y=2-x

-Calculate the volume using a double integral namely

http://img145.imageshack.us/img145/2705/q3tv3.png [Broken]
3. The attempt at a solution

For the first part I have firstly drawn the four lines in the xy plane
http://img239.imageshack.us/img239/7389/q3ct1.png [Broken]

Obviously I make it 3d by adding the top (i.e. 1+x^2+y^2 for -1 ≤ x ≤1 and 0 ≤ y ≤ 2)

My question is regarding the volume integral. It is difficult to nominate the y bounds due to the nature of the 4 lines. I am thinking I should split the area into two sections, bisected by the y axis such that the volume can be calculated by:

http://img73.imageshack.us/img73/8666/q3ao8.png [Broken]

Am I on the right track?

Last edited by a moderator: May 3, 2017
2. Oct 6, 2008

### gabbagabbahey

Looks like a good plan to me!

3. Oct 6, 2008

### t_n_p

hmmm, noticed that I need to reverse the order of integration (noting also I must change the terminals) though as before I was ending up with x's after the second integration. That is, I need to perform integration with respect to y then with respect to x such that

http://img98.imageshack.us/img98/5985/q3ao8iu3.png [Broken]

Last edited by a moderator: May 3, 2017
4. Oct 6, 2008

### gabbagabbahey

Yes, I thought your original order of integration was a typo.

5. Oct 6, 2008

### t_n_p

also noticed my terminals are wrong!

for the first part with respect to y I should be integrating from x to -x+2 and for the second part I should be integrating from -x to x+2.

After fixing it up I got a volume of 14/3 units cubed. Is there any way to check this answer?

Last edited: Oct 6, 2008