Question

Determine the integral f(x,y)dA where f(x,y)=e^(-x^2-y^2) and R is the region satisfying x>=0, y>=0, x^2+y^2<=4, and x^2+y^2>=1 Hint: use polar integration and describe the region R in terms of polar coordinates

Answer 1

So first we setup the diffeomorphism between a rectangle and the annulus.
$$g(r, \theta) = (r\cos \theta, r \sin \theta)$$
Now, tell me the rectangle whose image is the required set \(\{1 \leq x^2 + y^2 \leq 4\}\)

Answer 2

So the restriction gives us information about the inner radius and outer radius of the annulus.

Answer 3

Do you understand what we are trying to do right now? We want to find the new domain we will be using after applying the polar coordinate transformation.

Answer 4

Well this is the annulus we are looking at.

Answer 5

ok. so the rectangle [1,4] ?

Answer 6

That's not a rectangle. Think carefully we are working in polar coordinates.

Answer 7

It's \([1, 4] \times [0, 2\pi]\)

Answer 8

ok

Answer 9

Ok yea, all of the A's should say B correct? can you continue a bit more? Sorry.. I'm trying to understand this.

Answer 10

Wait, you better double check that, there might be a mistake at the end.

Answer 11

which part?

Answer 12

Is 3\[\pi\] R in terms of polar coordinates?

Answer 13

and that is polar coordinate terms?

Answer 14

The integral is the same even though we applied the polar coordinate transform. Read about the change of variables theorem.

Answer 15

ok thank you so much!

Answer 16

One more correction, my apologies.

So let \(A=(1,4)×(0,2\pi)\) and apply the change of variables with polar coordinate transform \(g(r,\theta)=(r \cos \theta,r \sin \theta)\) and we get
$$
\newcommand{\e}{\mathrm{e}}
\begin{eqnarray*}
&&\int_{R}\e^{-(x^2 + y^2)} \\
&=& \int_A\e^{-((r \cos \theta)^2 + (r \sin \theta)^2)}|r| \\
&=& \int_A\e^{-r^2((\cos \theta)^2 + (\sin \theta)^2)}r \\
&=& \int_{(1, 4) \times (0, 2\pi)}\e^{-r^2}r \\
&=& \int_0^{2\pi}\int_1^4\e^{-r^2}r \\
&=& 2\pi\int_1^4\e^{-r^2}r \\
&=& 2\pi\int_1^4\e^{u}r \cdot \frac{-1}{2r}\\
&=& -\pi\int_1^4\e^{u} \\
&=& -\pi \cdot (-e+e^4) \\
&=& (\e-\e^4)\pi
\end{eqnarray*}
$$