# Multivariable caculus: Am I or is Maple wrong?

Well, the problem goes as follows:

$$\int_{\Omega}(x^2+y^2)dxdy\quad\Omega:=\{(x,y)\mid x^4+y^4\le 1\}$$

My approach:
By symmetry I needed only to find $$\int_{\Omega_1}(x^2+y^2)dxdy\quad\Omega_1:=\{(x,y)\mid x^4+y^4\le 1\wedge x,y\ge 0\}$$ Let $(x,y)=(r\cos\theta, r\sin\theta)$ where $\theta\in[0,\frac{\pi}{2}]$ and $r\in[0,(\cos^4\theta+\sin^4\theta)^{-1/4}]$ (the constraint for $r$ follows naturally from the boundary equation, I think). Thus the transformation Jacobi determinant is $\partial(x,y)/\partial(r,\theta)=r$, and $$\int_{\Omega_1}(x^2+y^2)dxdy=\int_{D_{r\theta}}r^2\cdot rdrd\theta\quad D_{r\theta}:=\{(r,\theta)\mid \theta\in[0,\frac{\pi}{2}]\wedge r\in[0,(\cos^4\theta+\sin^4\theta)^{-1/4}]\}$$ After some simple adjustment I got $$\int_{D_{r\theta}}r^2\cdot rdrd\theta=\frac14\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sin^4\theta+\cos^4\theta}=\frac14\frac{\pi}{\sqrt 2}$$ which was also confirmed by Maple. Therefore $$\int_{\Omega}(x^2+y^2)dxdy=\frac{\pi}{\sqrt 2}$$
However, when I switch to another approach: directly applying Fubini's Theorem, the result seems to be different.

By Fubini, we immediately have $$\int_{\Omega_1}(x^2+y^2)dxdy=\int_{0}^{1}dx\int_{0}^{(1-x^4)^{1/4}}(x^2+y^2)dy=\int_{0}^{1}(x^2\cdot(1-x^4)^\frac14+\frac13(1-x^4)^\frac34)dx$$ This integral was undoable by hand. So I turned to Maple for a numerical result just to see if my previous result is right, and Maple gave me this $$\int_{0}^{1}(x^2\cdot(1-x^4)^\frac14+\frac13(1-x^4)^\frac34)dx\approx0.9443468503$$ But $\frac14\frac{\pi}{\sqrt 2}\approx 0.55536$. It was quite disappointing. I went through my previous approach one more time and still couldn't find out where I was wrong. I think maybe I'm right, but I have always trusted Maple...

Smart people of MSE, could you help me out? I'd be grateful to any kind of clarification.

EDIT I was using the "evalf" function, (BTW I'm a newbie at Maple and literally don't know how to do multivariable calculus on that)

• Not sure what you did, but I am getting the same answer both ways. – Braindead Jun 18 '15 at 15:08
• I think the problem might lie in the transformation, but I can't prove that it is NOT a diffeomorphism between adjusted $\Omega_{1}$ and $D_{r\theta}$ (both shorn of a Lesbegue measure zero set and left to be two open sets so that transformation is a diffeomorphism) – Vim Jun 18 '15 at 15:08
• 4*integrate(x^2+y^2,x=0..surd(1-y^4,4),y=0..1) on maple gives me $\dfrac{\pi}{\sqrt{2}}$. – Braindead Jun 18 '15 at 15:09
• Yup. Both polar and Fubini gives me the same $\pi/\sqrt{2}$. – Braindead Jun 18 '15 at 15:10
• Holy.. I just realized that I should insert $\cdot$ between $x^2$ and $(1-x^4)^{1/4}$ in my first integrand term.. – Vim Jun 18 '15 at 15:21
There's something wrong with the numerical value of the second. Let us do the integral exactly. First, I agree with your form $$\int_0^1 \left( x^2 (1-x^4)^{1/4} + \frac{1}{3}(1-x^4)^{3/4} \right) \, dx.$$ Let's change variables to $x=u^{1/4}$ to start with. Then we have $dx = \frac{1}{4} u^{-3/4} \, du$, so the integral becomes $$\frac{1}{4}\int_0^1 \left( u^{-1/4} (1-u)^{1/4} + \frac{1}{3}u^{-3/4}(1-u)^{3/4} \right) \, du.$$ We now have two integrals of the form $\int_0^1 u^{-s}(1-u)^s \, du = \int_0^1 (u^{-1}-1)^s \, du$, so let's try and evaluate a general one, and hence only do one calculation. Set $u= \frac{1}{1+v}$, so $du = -\frac{dv}{(1+v)^2},$ $$\int_0^1 (u^{-1}-1)^s \, du = \int_0^{\infty} \frac{v^{s}}{(1+v)^2} \, dv.$$ Integrating this by parts gives $$\int_0^{\infty} \frac{v^{s}}{(1+v)^2} \, dv = 0 + s\int_0^{\infty} \frac{v^{s-1}}{1+v} \, dv,$$ providing $0<s<1$. I've done this integral before: is $$\pi s\csc{\pi s},$$ so we find that the overall answer is $$\frac{\pi}{4} \left( \frac{1}{4\sin{(\pi/4)}}+\frac{1}{3}\frac{3}{4\sin{(3\pi/4)}} \right) = \frac{\pi}{4\sqrt{2}}.$$