Average Temperature of Elliptical Plate

Take the surface average to find the answer. $\displaystyle \frac{1}{8\pi}\iint_{S}\left(5xy+x^{2}\right)\,dx\,dy$

Where $S$ is the region inside the ellipse. Make a variable change of $\displaystyle x=2u\,\,\,y=4v$ to contract the ellispe into a circle. The absolute value of the Jacobian of transformation is $8$

Now the integral becomes

$\displaystyle \frac{1}{\pi}\iint_{T}\left(40uv+4u^{2}\right)\,du\,dv$

Where $T$ is the region inside the circle $u^{2}+v^{2}=1$ in the $uv$ plane.

Changing to polar coordinates we have

$\displaystyle \frac{1}{\pi}\int_{0}^{2\pi}\int_{0}^{1}\left(40r^{2}\sin(\theta)\cos(\theta)+4r^{2}\cos^{2}(\theta)\right)r\,\,dr\,d\theta$

It is easy to see that the integral $\displaystyle \int_{0}^{2\pi}\sin(\theta)\cos(\theta)\,d\theta = \int_{0}^{2\pi}\frac{1}{2}\sin(2\theta)\,d\theta=0$

And also $\displaystyle \int_{0}^{2\pi}\cos^{2}(\theta)\,d\theta = \pi$

Using the above results and solving the above double integral we arrive at the answer that the average temperature is $\large 1$