# Average temperature of a hotplate

1. Apr 18, 2010

### edough

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

An 8-inch hot plate is described by the region R = {(x,y) | x2 + y2 16}. The temperature at the point (x,y) is given by T(x,y) = 400 cos(0.1 ), measured in degrees Fahrenheit. What is the average temperature of the hot plate?

2. Relevant equations

would i use the centers of mass equations and then plug in the x, y values that i get into the Temperature equation? (Can you even centers of mass for a temp average?)
X= My / m and Y = Mx / m

or should i set up a double integral or something? im confused!

2. Apr 19, 2010

### Dickfore

no. it has to do with the stationary solution of the heat transfer equation:

$$\frac{\partial u}{\partial t} = a^{2} \nabla^{2} u,$$

which, under the assumption of stationarity $${\partial u}/{\partial t} = 0$$, reduces to the Laplace equation:

$$\nabla^{2} u = 0,$$

where u(x, y) is the stationary temperature distribution.

There are powerful theorems for elliptic partial differential equations that enable you to find the solution.

3. Apr 19, 2010

### Dick

Oh, come on. It's not a PDE problem. Just integrate T(x,y) over the plate and divide by the area of the plate. As read, the average temperature is 400*cos(0.1), it's a constant. There's clearly some formatting problems with the original question. The region of the plate should be x^2+y^2<=16.

4. Apr 19, 2010

### edough

I'm sorry. the original equation for T(x,y) = 400cos (0.1 sqrt(x^2 +y^2)) and yes the region is x^2 + y^2 <= 16. Very sorry about that..

So, I was trying to use a double integral with polar coordinates..

int (from 0 to 2pi) 2nd integral (from 0 to 4) [400cos( 0.1sqrt(r^r) )] *r dr d(theta)

Is this the right way of going about it?

5. Apr 19, 2010

### Dickfore

Yes, and you should also divide by the total area of the region.

6. Apr 19, 2010

### edough

Then How would i go about integrating 400cos(0.1*sqrt(r)) * r dr ??

7. Apr 19, 2010

### Dickfore

Make the substitution $$t = 0.1 \sqrt{r}$$

8. Apr 19, 2010

### Dick

400cos (0.1 sqrt(x^2 +y^2)) is 400*cos(0.1*r). There's no sqrt(r) here. Just integrate by parts.