Double integral computation

Question

Let

$$ S=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq4,\,x^2+y^2-4y\geq0,\,x\geq0,\,y\geq0\}. $$

I am then asked to evaluate $$\iint_S x\,e^{4y}\,dx\,dy.$$

The answer I got is $$-\frac{1}{8}(e^4-5).$$ Is it correct?

Answer 1

You're off by a sign, I think. Your answer must be positive. Your setup must have been very close. I have $$\int_0^1\int_{\sqrt{4-(y-2)^2}}^{\sqrt{4-y^2}}x\,e^{4y}\,dx\,dy=\frac{e^4-5}{8}.$$

Answer 2

I had to break the integral into two.

$\iint_{S}xe^{4y}dxdy = \int_{0}^{\sqrt{3}}xdx\int_{0}^{2-\sqrt{4-x^2}}e^{4y}dy + \int_{\sqrt{3}}^{2}xdx\int_{0}^{\sqrt{4-x^2}}e^{4y}dy$.

then I compute it and I got the answer: $\frac{e^4-5}{8}$

This is another way of integrate :)