The problem is this-

Evaluate $\int\int_R \sqrt{xy - y^2}dxdy$ where R is a triangle with vertices (0,0), (10,1) and (1,1).

I have done up to this- $\int_{y=0}^{1}\int_{x=y}^{10y} (x^2 - xy)^{\frac12}dxdy = \int_{y=0}^{1}dy\int_{x=y}^{10y}(x^2 - xy)^{\frac12}dx$

I was stuck here, and opened the book to check up the solution, and I found that it is done as $\int_{y=0}^{1}dy\int_{x=y}^{10y}(x^2 - xy)^{\frac12}dx = \int_{y=0}^{1}dy [\frac23 \frac1y (xy - y^2)^{\frac32}]^{10y}_{y}$
Now, I have got no idea where the $\frac1y$ came from. And how the step is done.
Can anyone please clarify?

  • 2
    $\begingroup$ $\frac1y$ comes from the fact that $y$ is like constant multiplied to $x$ in the inner integral. $\endgroup$ – Sonal_sqrt Feb 19 '18 at 8:40
  • $\begingroup$ Is it $xy-y^2$ or $x^2-xy$ under the square root? $\endgroup$ – Christian Blatter Feb 19 '18 at 10:14

I think solving it in polar coordinates is easier.$$I=\iint_R \sqrt{xy - y^2}dxdy=\iint_R \sqrt{r^2\sin\theta\cos\theta - r^2\sin^2\theta}rdrd\theta\\= \iint_R \sqrt{r^2\sin\theta\cos\theta - r^2\sin^2\theta}rdrd\theta\\=\iint_R r^2\sqrt{\sin\theta\cos\theta-\sin^2\theta}drd\theta\\=\int_{\tan^{-1}0.1}^{\frac{\pi}{4}}\int_{0}^{\frac{1}{\sin\theta}}r^2\sqrt{\sin\theta\cos\theta-\sin^2\theta}drd\theta\\=\dfrac{1}{3}\int_{\tan^{-1}0.1}^{\frac{\pi}{4}}(1+\cot^2\theta)\sqrt{\cot\theta-1}d\theta\\=-\dfrac{1}{3}\int_{\tan^{-1}0.1}^{\frac{\pi}{4}}\sqrt{\cot\theta-1}d(\cot\theta)\\=\dfrac{1}{3}\int_{1}^{10}\sqrt{u-1}du\\=\dfrac{2}{9}(u-1)^\frac{3}{2}|_{1}^{10}=6$$ Here is a sketch of region $R$:

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