# Double Integral (underneath a surface and above a square)

1. Jan 10, 2012

### letalea

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

The volume underneath the surface z= y/ (1+xy) and above the square {(x,y)| 2≤x≤3 , 3≤
y≤ 4} is:

2. Relevant equations

3. The attempt at a solution

My professor had provided us with 8 possible solutions (where only one is correct). However the answer I had produced did not match with any of the 8. Any insight to where/how I went wrong is greatly appreciated!

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2. Jan 10, 2012

### SammyS

Staff Emeritus
Your solution looked good to me.

What were the eight choices given?

There is a mistake in doing each of the last two integrals.

You should have coefficients of 1/3 & 1/2 respectively.

Last edited: Jan 10, 2012
3. Jan 11, 2012

### letalea

Could you explain this mistake to me? As I am not seeing how I come to get the coefficients.

But yes you are right, when you factor in the coefficients it does give me one of the possible answers the prof gave us!

4. Jan 11, 2012

### SammyS

Staff Emeritus
$\displaystyle \int\ln(1+3y)dy$

Let u = 1+3y → du = 3 dy → $\displaystyle dy=\frac{1}{3}du$

Can you take it from there?

5. Jan 11, 2012

### letalea

Yes, that makes perfect sense. Thank you!