Double Integral (underneath a surface and above a square)

[letalea]

Homework Statement

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

Homework Equations

Please see attachment.

The Attempt at a Solution

Please see attachment for 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!

Thanks in Advance!

 

[SammyS]

Homework Statement

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

Homework Equations

Please see attachment.

The Attempt at a Solution

Please see attachment for 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!

Thanks in Advance!

Your solution looked good to me.

What were the eight choices given?

Added in Edit:

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

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

 

[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!

 

[SammyS]

\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?

 

[letalea]

Yes, that makes perfect sense. Thank you!