# Triple Integral

• Jun 5th 2010, 07:48 PM
demode
Triple Integral
Here's a worked question:
http://img249.imageshack.us/img249/5356/90726987.gif

Shouldn't the second to last line be:

$-3 \int^2_0 \int^3_0 \int^1_0 u^2 du dv dw$

where -3 is the Jacobian? Where did the $\frac{1}{3}$ come from?
• Jun 5th 2010, 08:01 PM
undefined
Quote:

Originally Posted by demode
Here's a worked question:
Shouldn't the second to last line be:

$-3 \int^2_0 \int^3_0 \int^1_0 u^2 du dv dw$

where -3 is the Jacobian? Where did the $\frac{1}{3}$ come from?

Haven't done these recently enough to fully check all the steps (without spending a lot of time brushing up), however I can point out that (in the image) the third line up from the bottom has the absolute value of the reciprocal of the expression that came out to -3 above. So the absolute value of the reciprocal of -3 is 1/3. If that's not enough then I'm sure someone else will be able to help you further.
• Jun 5th 2010, 09:59 PM
AllanCuz
Quote:

Originally Posted by demode
Here's a worked question:
http://img249.imageshack.us/img249/5356/90726987.gif

Shouldn't the second to last line be:

$-3 \int^2_0 \int^3_0 \int^1_0 u^2 du dv dw$

where -3 is the Jacobian? Where did the $\frac{1}{3}$ come from?

It came from the change of bounds. If you go through all the steps, and note the difference in the bounds from before changing to the jacobian to after, you will note that the value makes sense.