# Integrate the function

• March 21st 2009, 11:25 AM
althaemenes
Integrate the function
Hi, I am kinda having trouble trying to find the limits for the triple integral.

Thanks.

Integrate the function https://instruct.math.lsa.umich.edu/...b7704d4671.png over the solid given by the figure below, if P=(-4,-4,2), Q=(-3,-3,2), R=(3,3,2), and S=(4,4,2).
integral =
(You should find the actual value when you evaluate the integral!)
• March 21st 2009, 12:45 PM
HallsofIvy
Are we to assume the bottom is in the xy-plane? Use cylindrical coordinates, of course.
$\theta$ goes from $\pi/4$ to $5\pi/4$ (arctan(4/4)= arctan(1) to arctan(-4/4)), r from $3\sqrt{2}$ to $4\sqrt{2}$, and z from 0 to 2.
• March 21st 2009, 02:37 PM
althaemenes
Thanks a lot for the reply(Clapping)...

I tried to integrate the function and getting 2960/3 which is not the right answer...

okay.. first I converted to cylindrical coordinate!!!

Given, -5x+5y

so -5(r*cos(t))+5(r*sin(t))

now integrated the following function:

=> (-5(r*cos(t))+5(r*sin(t)))*r*dz*dr*dt

=> (740/3)*sqrt(2)*sin((1/4)*Pi)+(740/3)*sqrt(2)*cos((1/4)*Pi)-(740/3)*sqrt(2)*sin((5/4)

=> 2960/3 (Doh)(Doh)(Doh)
• March 21st 2009, 03:20 PM
althaemenes
finally..
the answer is -2960/3 ... so I guess its not a volume...(Bow)

Thanks again....
• March 21st 2009, 05:04 PM
HallsofIvy
what would make you think it was a volume?
• March 24th 2009, 07:32 AM
althaemenes
volume!!!
Well, double and triple integral could be volume if( f(x)>=0), so I assumed that they wanted me to find the volume... but I was wrong... (Crying)