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I'm given the following exercise:

$\iint\limits_D \exp(x^{2}+y^{2})dA$

And I dont even know where to start, any chance someone could give me a hint?

D is a half circle, given by:

$9\le x^{2}+y^{2}\le 16, y\ge 0$

Wheres A, I don't know what it is, its not given in the exercise. - Thats partly whats bugging me.

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What are $D$ and $A$ supposed to be? –  J. M. Dec 2 '10 at 17:33
(on a bit of limb) how about starting with $dA=dxdy$ ? –  Will Robertson Dec 2 '10 at 17:33
What's $D$? If it's the unit disc you can use polar coordinates. Then you will get $$2\pi\int_0^1 r\cdot\mathrm{exp}(r^2)dr=\pi(\mathrm{exp}(1)-\mathrm{exp}(0))=\pi(e-1).$$ –  Rasmus Dec 2 '10 at 17:35
So you have half an annulus... try determining the solution for a general radius $r$, call it $I(r)$, and try computing $I(4)-I(3)$. –  J. M. Dec 2 '10 at 17:46
For $A$ ... could you try looking at the associated chapter(s) and see how it was previously used? –  J. M. Dec 2 '10 at 17:48

3 Answers 3

up vote 5 down vote accepted

The notation $\underset{D}{\displaystyle\iint }f(x,y)\;\mathrm{d}A$ has the same meaning as $\underset{D}{\displaystyle\iint }f(x,y)\;\mathrm{d}x\mathrm{d}y$.

To evaluate an integral in a different coordinate system one has to find the absolute value of the Jacobian determinant of the transformation. Using polar coordinates, we can transform the initial integral

$$\underset{D}{ \iint }e^{x^{2}+y^{2}}\;\mathrm{d}A=\underset{D}{\iint }e^{x^{2}+y^{2}}\;\mathrm{d}xdy$$

into this one

$$\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) \mathrm{d}\theta ,$$

where the conversion factor $r$ is the Jacobian determinant (see Example 3 on the Wikipedia article), and observe that $r^{2}=x^{2}+y^{2}$:

$$\begin{eqnarray*} \underset{D}{\iint }e^{x^{2}+y^{2}}\mathrm{d}A &=&\underset{D}{\iint } e^{x^{2}+y^{2}}\mathrm{d}x\mathrm{d}y \\ &=&\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) \mathrm{d}\theta \\ &=&\int_{0}^{\pi }\left[ \frac{1}{2}e^{r^{2}}\right] _{3}^{4}\mathrm{d}\theta \\ &=&\int_{0}^{\pi }\frac{1}{2}(e^{16}-e^{9})\mathrm{d}\theta \\ &=&\frac{\pi }{2}(e^{16}-e^{9}). \end{eqnarray*}$$

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The A in $dA$ refers to the infinitesimal thing you're integrating over. It just means that instead of a line, you integrate over a surface. A for Area.

As for a hint to start: Since you are dealing with circles, polar coordinates are a natural choice. I am sure these coordinates have been discussed in your lecture before. Note that $x^2 + y^2 = r^2$, where $r$ is the distance of the point from the coordinate origin. Also note that $dA = r dr d\phi$

Then, you can characterize the area $D$ by simple intervals for your polar coordinates: r will go from $3$ to $4$ and $\phi$ will go from $0$ to $\pi$.

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Okay, I've gotten the 3 and 4, with $0$ to $\pi$ already, in the previous exercise. - and we've been around polarcoordinates yes, or a little. - its just extra homework I'm doing :) - I dont however get that; $dA = rdrd\theta$ part of what your saying. –  Emil Madsen Dec 2 '10 at 18:04
Okay. $dA$ is what you get when each of the coordinates is changed infinitesimally. In cartesian coordinates, that is simply the rectangle $dx dy$. In polar coordinates, draw a line of radius r. Then add $dr$ to the radius and change the angle by $d\phi$. This gives a shape that isn't really rectangular, but for infinitesimal changes it doesn't matter. Now, one side of it has length $dr$, the other side has length $r d\phi$. Thus, the area of the shape is $r dr d\phi$. –  Lagerbaer Dec 2 '10 at 18:09

Some confusion is coming from not understanding what the $dA$ represents, so here's a link on the area element in polar coordinates that hopefully will help clear up this point.

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