Triple Integral to Find Avg Temperature

[heartshapes]

I have tried this problem at least 20 times and get lost of the integration. I think I am making it too complex or I am just really confused.

Temperature of a solid is : T= f(x,y,z) = 3x-2y+z+30

The integral I made is:

\(\displaystyle \int_{0}^{\frac{30}{2}} \int_{0}^{\frac{2y-30}{3}} \int_{0}^{2y-3x-30} dzdxdy\)

Is this right? After i do the integration I keep ending up with numbers in the 5000's and other huge numbers. when like f(4,6,2) is 32 so it makes no sense.

THANKS!

 

[galactus]

What is the exact wording of the problem to make sure we have the correct integral set up.

Evaluating the integral at hand, I get 750.

 

[heartshapes]

What is the exact wording of the problem to make sure we have the correct integral set up.

Evaluating the integral at hand, I get 750.

use a triple integral to determine the avg temperature of the material in the solid box. show all units for all quantities in your final answer.

 

[Deleted member 4993]

I have tried this problem at least 20 times and get lost of the integration. I think I am making it too complex or I am just really confused.

Temperature of a solid is : T= f(x,y,z) = 3x-2y+z+30

use a triple integral to determine the avg temperature of the material in the solid box. show all units for all quantities in your final answer.

The integral I made is:

\(\displaystyle \int_{0}^{\frac{30}{2}} \int_{0}^{\frac{2y-30}{3}} \int_{0}^{2y-3x-30} dzdxdy\)

Is this right? After i do the integration I keep ending up with numbers in the 5000's and other huge numbers. when like f(4,6,2) is 32 so it makes no sense.

THANKS!

Let the dimension of the box be [a,b,c] and (0,0,0) be one corner.

Then the equation for average temperature would be

\(\displaystyle T_{avg} \, = \frac {1}{abc} \cdot \int_{0}^{b}\int_{0}^{a} \int_{0}^{c} (2y-3x-30)dzdxdy\)