Integration of Bounded Region I'm trying to solve this problem:
The region of integration is the triangle $D$ with vertexes $A(0,0),B(1,1),C(10,1)$.
Find the solution of $\iint_D \sqrt{x^2-y^2}\,dx\,dy$.
MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\}$.
After that the integral becomes:
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy$$
We can solve first this indefinite integral $$\int\sqrt{x^2-y^2}\,dx$$
where $y\in[0,1]$ is a constant.
So$$\begin{align}\int\sqrt{x^2-y^2}\,dx=x\sqrt{x^2-y^2}-\int\frac{x^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\frac{x^2-y^2+y^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{1}{\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{(1+\frac{x}{\sqrt{x^2-y^2}})}{x+\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\log (x+\sqrt{x^2-y^2})\end{align}$$
The result is:
$$\int\sqrt{x^2-y^2}\,dx=\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})+C$$
We can now solve the definite integral
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy=\int_0^1\Bigl(\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})\Bigr)\Big|_y^{10y}\,dy\\=\int_0^1 \Bigl(5y^2\sqrt{99}-\frac{1}{2}y^2\log (10+\sqrt{99})\Bigr)\,dy\\=\frac{5\sqrt{99}}{3}-\frac{\log (10+\sqrt{99})}{6}
$$
 A: Riemann integration for the area integrals is nasty as you need to find a way to enumerate the whole area.
I would suggest slightly different way, i.e. randomly picking a point from the triangle:
$$ f(x,y) = \sqrt{x^2-y^2} $$
$$ \int_D {f(x,y)} = A \cdot \frac{\sum_{n=1}^{k}{f(x_n,y_n)}}{k}, (x_n,y_n)\in D, k>1200$$, where $(x_n,y_n)$ is chosen randomly from the triangle and A is the area of the triangle. This is called "average-based integral", and it is slightly easier to use for areas like D than riemann integral.
Picking a point from a triangle with nice distribution can be done by picking a point from bounding box, and then filtering out random points that are outside of the triangle, as in:
$$ \{ (x,y) \in [x_0..x_1]\times [y_0..y_1] | (x,y) \in D \} $$
A: Using your parametrization we have:
$$\begin{align}\iint_{D_1}\sqrt{x^2-y^2}dxdy&=\int_0^1dx\int_{y=\frac{1}{10}x}^{y=x}\sqrt{x^2-y^2}dy\\&=\int_0^1xdx\int_{y=\frac{1}{10}x}^{y=x}\sqrt{1-\left(\frac{y}{x}\right)^2}dy \:\: \blacktriangle\end{align}$$
Now, suppose we know that (there's pleny of proofs of this fact on the net): $$\int\sqrt{1-u^2}du=\frac{\arcsin(u)+u\sqrt{1-u^2}}{2}+C$$
So: $$\int\sqrt{1-\left(\frac{y}{x}\right)^2}dy=\frac{x\arcsin\left(\frac{y}{x}\right)+y\sqrt{1-\left(\frac{y}{x}\right)^2}}{2}+C \:\: \star$$
Combining $\blacktriangle$ and $\star$ together we obtain:
$$\begin{align}\iint_{D_1}\sqrt{x^2-y^2}&=\int_0^1x\left[\frac{x\arcsin\left(\frac{y}{x}\right)+y\sqrt{1-\left(\frac{y}{x}\right)^2}}{2}\right]_{y=\frac{1}{10}x}^{y=x}dx\\&=\frac{1}{6}\left[\frac{\pi}{2}-\left(\arcsin\left(\frac{1}{10}\right)+\frac{\sqrt{99}}{100}\right)\right]\end{align}$$
In a similar way you can calculate $\iint_{D_2}\sqrt{x^2-y^2}dxdy$.
As suggested, there are better ways to parametrize your domain of integration
A: MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\}$.
After that the integral becomes:
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy$$
We can solve first this indefinite integral $$\int\sqrt{x^2-y^2}\,dx$$
where $y\in[0,1]$ is a constant.
So$$\begin{align}\int\sqrt{x^2-y^2}\,dx=x\sqrt{x^2-y^2}-\int\frac{x^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\frac{x^2-y^2+y^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{1}{\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{(1+\frac{x}{\sqrt{x^2-y^2}})}{x+\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\log (x+\sqrt{x^2-y^2})\end{align}$$
The result is:
$$\int\sqrt{x^2-y^2}\,dx=\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})+C$$
We can now solve the definite integral
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy=\int_0^1\Bigl(\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})\Bigr)\Big|_y^{10y}\,dy\\=\int_0^1 \Bigl(5y^2\sqrt{99}-\frac{1}{2}y^2\log (10+\sqrt{99})\Bigr)\,dy\\=\frac{5\sqrt{99}}{3}-\frac{\log (10+\sqrt{99})}{6}
$$
