Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\int\int\int_{V}(x-y)dV$$ 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.

share|improve this question
add comment

3 Answers 3

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 $0$.

share|improve this answer
    
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
add comment

Cylindrical coordinates:

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

so

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

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

share|improve this answer
    
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
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.