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$$\int\int\int_{V}(x-y)dV$$ where $V$ is volume enclosed by : $$ S=\left\{(x,y,z):(x^{2}+y^{2})^{2}+z^{4}=16;z\geq0\right\}$$

What I did: $\int\int\int_{V}(x-y)dV=\int\int_{A}\left[\int_{0}^{\left(16-(x^{2}+y^{2})^{2}\right)^{1/4}}(x-y)dz\right]dA=\int\int_{A}(x-y)\left(16-(x^{2}+y^{2})^{2}\right)^{1/4}dA$ where $A=\{(x,y):x^{2}+y^{2}\leq4$ I tryed changing to polars next, but didn't helped much... I don't think it's relevant, but $(x-y)$ is a $div (f)$ that I got earlyer.

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up vote 1 down vote accepted

If you interchange the roles of $x$ and $y$, the domain $V$ does not change, but the integrand $x-y$ will become $y-x=-(x-y)$, which imply that the value of the integral must be $0$.

Remark: Indeed, a similar trick shows that both $\iiint_V xdV$ and $\iiint_V ydV$ are $0$.

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could you please explain this a bit more @richard – Mykolas Nov 17 '12 at 12:08
@Mykolas: If you consider an integral $\iiint_V f(x,y,z)d V$, and you know that $T:V\to V$, $T(x,y,z)=(y,x,z)$ is a bijection. Then you may consider $T$ as a change of variables, which gives you $\iiint_V f(x,y,z)d V=\iiint_V f\circ T(x,y,z)|\det T|d V=\iiint_V f(y,x,z)dV$. Letting $f(x,y,z)=x-y$, you will see the integral must be $0$. – 23rd Nov 17 '12 at 12:14

Cylindrical coordinates:

$$x=r\cos t\;\;,\;\;y=r\sin t\;\;,\;\;z=z\;\;\;,\;\;|J|=r\geq 0\,\,,\,\,0\leq t\leq 2\pi\;\;,\;\;0\leq z\leq 16-r^4$$


$$\int_0^{2\pi}\int_0^2\int_0^{16-r^4} r^2(\cos t-\sin t)dz\,dr\,dt=\int_0^{2\pi}(\cos t-\sin t)dt\int_0^2r^2(16-r^2)\,dr=$$

$$=\left.\left(\sin t+\cos t\right)\right|_0^{2\pi}\int_0^2r^2(16-r^2)\,dr=0\cdot\int_0^2r^2(16-r^2)\,dr=0$$

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Could you, please, explain how you got bounds for $z$ and got read of $\sqrt[4]$ @DonAntonio – Mykolas Nov 17 '12 at 12:14
never mind. I got it. Thankyou – Mykolas Nov 17 '12 at 12:15

Note that $$(x,y,z)\in V \implies(y,x,z)\in V,$$ hence$$\iiint_V(x-y) =\iiint_V x -\iiint_Vy =\iiint_V x -\iiint_Vx =0.$$

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