Let $R$ a region defined by the interior of the circle $x^2+y^2=1$ and the exterior of the circle $x^2+y^2=2y$ and $x\geq 0$, $y\geq 0$

Using polar coordinates $x=r\cos t$, $y=r\sin t$ to determine the region $D$ in $rt$ plane that corresponds to $R$ under this change of coordinate system (polar coordinates)

i.e. since $T(r,t)=(r\cos t, r\sin t$)

How is $D$ such that $T(D)=R$

Calculate the integral $\int\int_Rxe^y \ dx \ dy$ using this change coordinate system (polar coordinates)

Attempt of solution

Region $R$ is in the interior of circle $x^2+y^2=1$ and the exterior of the circle $x^2+(y-1)^2=1$, like in next image Region R

Then I find region $D$ by sustituing polar coordinates in equations that restricts $R$:

$$x^2+y^2=1 \Rightarrow r=1$$

$$x^2+(y-1)^2=1 \Rightarrow r=2\sin(t)$$

Then, $$2\sin(t) \leq r\leq 1$$

Also, $$r\cos(t) \geq 0 , r\sin(t)\geq 0$$

Region D is shown in next image Region D

To find limits for $t$ I determined intersection of $r=2\sin(t)$ and $r=1$, which is when I clear $t=\arcsin(\frac{r}{2})$, so, because $r=1$, we have $t=\arcsin(\frac{1}{2})=\frac{\pi}{6}$, so intersection occurs in $(1,\pi/6)$

And $$0\leq t \leq \frac{\pi}{6}$$

Next I tried to solve the integral:

$$\int\int_Rxe^y \ dx \ dy=\int_0^{\pi/6}\int_{2\sin(t)}^1r\cos(t) \ e^{r\sin(t)} \ dr \ dt$$

And I get a lot of trouble when I trie to integrate respect to $t$

After integrated respect to $r$ I get:

$$\int\int_Rxe^y \ dx \ dy=\int_0^{\pi/6}\left(\frac{\cos t}{\sin t}e^{\sin t}-\frac{\cos t}{\sin^2 t}e^{\sin t}+\frac{\cos t}{\sin^2 t}e^{2\sin^2 t}-2\cos t \ e^{2\sin^2 t}\right) \ dt$$

But I don't kown how to proceed with this integral, I find that $\int \frac{e^x}{x}$ cannot be integrated

http:// math.stackexchange.com/questions/251795/problem-when-integrating-ex-x

https:// en.wikipedia.org/wiki/Exponential_integral

Also, I'm not sure if my region $D$ is correct

Any help will be appreciated


1 Answer 1


Two remarks:

  1. You forgot the Jacobian determinant when did the variable change. It should be extra $r$ in the integral, which makes the inner intergation w.r.t. $r$ even less attractive (integration by parts two times, brrr...).
  2. Try to switch the order of integration. As you say $t=\arcsin(r/2)$, so it becomes $$ \int_0^1\int_0^{\arcsin(r/2)}r\cos(t)e^{r\sin(t)}\cdot r\,dtdr=\int_0^1 \left[e^{r\sin(t)}\right]_0^{\arcsin(r/2)}r\,dr=\int_0^1r(e^{r^2/2}-1)\,dr. $$ It looks much nicer.

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