Double integration using change of variables, given the vertices of a triangle

Question

I am having problem solving this problem Compute $$\iint_\Omega xy \,dx\, dy\,,$$ where $\Omega$ is a triangle with vertices $$(0,0), (\cos \alpha, \sin \alpha),(-\sin \alpha, \cos \alpha).$$

I learned how to solve this type of problem when the vertices are given as numbers instead of trigonometric functions. But as this question has given in trigonometric form, I do not understand what to substitute with. Any type of hint will be helpful.

Answer 1

Hints: Both your points $A=(\cos \alpha, \sin \alpha)$ and $B=(-\sin \alpha, \cos \alpha)$ lie on unit circle, and it is easy to see that point $A$ lies on a line $y=\tan \alpha \cdot x$ which passing the origin, and that point $B$ lies on a line $y = - \cot \alpha \cdot x$ which also passing the origin $O$. Next, we have to find the equation for the line which passing our points $A$ and $B$, it is: $$y=\frac{\cos \alpha - \sin \alpha}{-\sin \alpha - \cos \alpha}\left( x - \cos \alpha \right) + \sin \alpha.$$

Now we can split the area of our triangle $OAB$ into two parts. The 'left' part will have these limits: $$-\sin \alpha \le x \le 0, \quad - \cot \alpha \cdot x \le y \le \frac{\cos \alpha - \sin \alpha}{-\sin \alpha - \cos \alpha}\left( x - \cos \alpha \right) + \sin \alpha,$$

and the 'right' part: $$0 \le x \le \cos \alpha, \quad \tan \alpha \cdot x \le y \le \frac{\cos \alpha - \sin \alpha}{-\sin \alpha - \cos \alpha}\left( x - \cos \alpha \right) + \sin \alpha$$

EDIT: In my hint $\alpha$ is sharp angle.

Figure.

Answer 2

We use first a linear change of coordinates with determinant one, $$ \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix}\ , $$ $$ \begin{bmatrix} u\\ v \end{bmatrix} = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}\ , $$ so that $\Omega$ is transformed to the triangle $\Delta$ with fixed vertices $(0,0)$, $(1,0)$, $(0,1)$. Then we apply Fubini. The computation is: $$ \begin{aligned} \iint_\Omega xy \;dx\; dy &= \iint_\Delta (\cos\alpha \cdot u -\sin \alpha \cdot v) (\sin\alpha \cdot u +\cos \alpha \cdot v) \;du\; dv \\ &= \cos\alpha\sin\alpha \underbrace{\left(\iint_\Delta u^2\;du\; dv\right)}_{=1/12} - \cos\alpha\sin\alpha \underbrace{\left(\iint_\Delta v^2\;du\; dv\right)}_{=1/12} \\ &\qquad + (\cos^2\alpha -\sin^2 \alpha) \underbrace{\left(\iint_\Delta uv\;du\; dv\right)}_{=1/24} \\ &= \frac 1{24}\cos(2\alpha)\ . \end{aligned} $$

Answer 3

In polar coordinates, the triangle area is enclosed by the three sides

$$\theta = \alpha,\>\>\>\theta =\frac\pi2+\alpha, \>\>\>r= \frac1{\sin(\theta-\alpha)+ \cos(\theta-\alpha)}$$ which are determined by the three vertexes $(0,0)$, $(\cos \alpha, \sin \alpha)$ and $(-\sin \alpha, \cos \alpha)$. The integral is then

\begin{align} \iint_\Omega xy \,dx\, dy &= \int_\alpha^{\frac\pi2+\alpha}(r^2\sin\theta\cos\theta)rdrd\theta \\ &=\frac18\int_\alpha^{\frac\pi2+\alpha} \frac{\sin2\theta d\theta}{[\sin(\theta-\alpha)+ \cos(\theta-\alpha)]^4} \\ &=\frac18\int_0^{\frac\pi2} \frac{\sin(2t+2\alpha)dt}{(\sin t+ \cos t)^4} \\ & =\frac18\int_0^{\frac\pi2} \frac{\sin2t\cos2\alpha+ \cos2t\sin2\alpha}{(1+\sin2t)^2}dt\\ & =\frac{\cos2\alpha}8\int_0^{\frac\pi2} \frac{\sin2t}{\ (1+\sin2t)^2}dt = \frac{\cos2\alpha}{24} \end{align}