# double integral over general region

• Nov 14th 2009, 12:30 PM
zpwnchen
double integral over general region
Quote:

Using the maxima and minima of the function, produce upper and lower estimates of the integral
http://webwork.asu.edu/webwork2_file...070d87fc21.png where http://webwork.asu.edu/webwork2_file...f15439ff51.png is the circular disk: http://webwork.asu.edu/webwork2_file...853ff30bc1.png.
??? http://webwork.asu.edu/webwork2_file...ddd8d53c41.png ???

how to do that?
• Nov 14th 2009, 04:07 PM
TKHunny
Maybe...

$\displaystyle e^{5 \cdot (0)} \le e^{5 \cdot (x^{2}+y^{2})} \le e^{5 \cdot (9)}$
• Nov 14th 2009, 04:12 PM
zpwnchen
i try to enter those in the answer box, but turn out that it's not right.
I try integrate those and i got : $\displaystyle 2e^{45}\pi -2\pi$ which is not right as well. :(
• Nov 14th 2009, 07:15 PM
TKHunny
Perhaps you should try those integrals again.

For the lower bound, I get $\displaystyle \frac{9\pi}{4}$.
• Nov 14th 2009, 07:19 PM
zpwnchen
$\displaystyle \frac{9\pi}{4}$ which is not correct either. :(
• Nov 14th 2009, 07:46 PM
TKHunny
Whoops. I did only one quadrant.

The question states, "produce upper and lower estimates". $\displaystyle 9\pi$ IS a lower bound.

Did you do as I asked? Did you try those integrals again or did you just type in my answer?
• Nov 14th 2009, 07:53 PM
zpwnchen
I used polar coordinate to estimate. For the lower bound, I try to integrate
r from 0 to 3 and $\displaystyle \theta$ from 0 to $\displaystyle \pi$

the answer turn out to be $\displaystyle e^{45} \pi - \pi$ and it is not correct.

In fact, i do not quite understand by "upper and lower estimates of the integral".
• Nov 15th 2009, 02:37 AM
TKHunny
1) Why [0,pi]? Try [0,2pi]

2) Why are you trying to evaluate the integral directly? That is not what is asked.

3) My first response entirely exposes the concept of bounds. We replace the argument with something simpler and see where it leads.