question :
find the value of $\iint_D \frac{x}{(x^2 + y^2)}dxdy$ ,
domain : $0\le x \le 1, x^2 \le y \le x$
,
so here, i tried to draw it first and i got that the domain is region in first quadrant bounded by $y=x^2$ and $y=x$ and i decided to convert the equation to polar coordinate,
\begin{align}
\int_{0}^{\pi/2}\int_{(r^2cos^2\theta)}^{rcos \theta}cos \theta dr d\theta\\
\end{align}
is this true?
i proceed to solve the integral, but still got r in the last equation. the answer of the book said $\frac{1}{2} log 2$
thankyou
$$\iint_D \frac{x}{(x^2 + y^2)}dxdy$$ ,
domain : $0\le x \le 1, x^2 \le y \le x$ is equivalent to implies $0\le y \le 1,$ $y \le x \le \sqrt{y} $ and $$\frac{\partial}{\partial x}\ln(x^2+y^2)= \frac{2x}{(x^2 + y^2)}$$
Therefore,
$$I=\iint_D \frac{x}{(x^2 + y^2)}dxdy = \int_{0}^{1}\int_{y}^{\sqrt{y}}\frac{x}{(x^2 + y^2)}dxdy = \int_{0}^{1}\frac{1}{2}[\ln((x^2 + y^2)]_y^{\sqrt{y}}dy =\int_{0}^{1}\ln(y+y^2)-\ln(2y^2)dy $$ ,
But $(x\ln x-x)' = \ln x$ and $$\ln(y+y^2)-\ln(2y^2) = -\ln y+\ln(1+y) -\ln2$$ Hence $$I= \frac{1}{2}\int_{0}^{1}\ln(y+y^2)-\ln(2y^2)dy =[(1+y)\ln(1+y)-(1+y) -y\ln y+y -y\ln 2 ]_0^1 = \frac{1}{2}\ln 2$$