I have calculus exercise to calculate double integral of function $f(x)=x^2+y^2$ over the area enclosed inside curve $x^4+y^4=1$. I have tried with polar coordinates:
$$
\iint_D f(\phi,r)r \,d \phi\,dr = 4\int_{0}^{\pi/2} \int_{0}^{[1/(\cos^4\phi+\sin^4\phi)]^{1/4}} r^3 d\phi dr
$$
Although this seems to be the right way (I get the right result with Wolfram Mathematica), it leads to the integral
$$
\int_{0}^{\pi/2}\frac{dx}{\cos^4x+\sin^4x}
$$
which I don't know how to easily execute.
I was wondering if there is any trick to use diferent new coordinate systems or integration by substitution? Is there any general trick to integrate a function ower the area of this curve, because it appears quite often in exercises?
Best Answer
From symmetry considerations, we can write $$\begin{align} \iint_{x^4+y^4\le 1}(x^2+y^2)\,dx\,dy&=8\int_0^1 \int_0^{(1-y^4)^{1/4}}x^2\,dx\,dy\\\\ &=\frac83\int_0^1 (1-y^4)^{3/4}\,dy\\\\ &=\frac23\int_0^1 (1-t)^{3/4}t^{-3/4}\,dt \tag1\\\\ &=\frac23 B\left(\frac74,\frac14\right)\tag2\\\\ &=\frac23 \Gamma(7/4)\Gamma(1/4)\tag3\\\\ &=\frac12\Gamma(3/4)\Gamma(1-3/4)\tag4\\\\ &=\frac12\frac\pi{\sin(3\pi/4)}\tag5\\\\ &=\frac\pi{\sqrt 2} \end{align}$$
NOTES:
In arriving at $(1)$, we made the substitution $y=t^{1/4}$.
In arriving at $(2)$ we recognized $(1)$ as the "standard" integral representation of the Beta function.
In going from $(2)$ to $(3)$ we used the relationship $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ between the Beta and Gamma functions.
In going from $(3)$ to $(4)$ we made use of the functional equation $\Gamma(1+x)=x\Gamma(x)$.
In going from $(4)$ to $(5)$ we used Euler's Reflection formula $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$