Does multivariate integral make sense and can it be evaluated?

Question

I have very hard instructor for multivariate calculus. He ask if the next integral is well-defined.

$$ \iint\limits_D\ {\cos(z)\sin^3(z)\cos(y)\sin(y)\over (\cos^2(z)+\sin^2(z)\cos^2(y))(\cos^2(z)+\sin^2(z)\cos^2(y)-b)}\,dy\,dz $$

$D$ is region $[0,\pi] \times [0,\pi]$, and $b$ live in $(0,1]$. Is it possible to calculate integration with Mathematica or by hand for all $b$? I consider this as improper integral. For b=1, free online version of Mathematica says integral is 0, but is not strong enough for other b.

Answer 1

The integral is always zero.

To see this, preform the replacement: $$u = -\cos(y), \ v = -\cos(z) \ \rightarrow du = \sin(y)dy, \ dv = \sin(z)dz$$ The integral is then: $$I = \iint\limits_D\ {uv(1-v^2) \over (v^2+(1-v^2)u^2) (v^2+(1-v^2)u^2 - b)}\,du\,dv $$ Where $D = [-1,1]\times[-1,1]$. Note that the integral is odd in both $u$ and $v$, leading to $I = 0\ $ for all $b$.

Now, how do we know this converges? the left hand denominator is never zero, but the right side will always be zero for some pair of values $u_0,v_0$. You can easily see that the roots lie on a closed symmetrical curve, i.e. for any value $b$, if $u_0,v_0$ is a root so are $\pm u_0,\pm v_0$. This means you can divide the integral into a region outside this "ring of roots" and one inside. Due to symmetry and the oddness, each of these two integrals converges to zero, and so does the sum.