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$F=xi+y^2j+(z+y)k$ then $S$ is boundary $x^2+y^2=4$ between the planes $z=x$ and $z=8$. Verify Divergence Theorem

I'm trying to verify the Divergence theorem, but I'm not sure of the results. I found the volume but i think it is wrong. I can't find the flux on the surfaces. Thank you very much for any help.

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    $\begingroup$ What did you try? $\endgroup$ – DiegoMath May 16 at 17:25
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    $\begingroup$ Let's go step by step. I tried to find the volume before. DivF=2y+2 and integration limits ; z from x to 8. then x from 0 to (4-y^2)^0.5 and y from -1 to 1 . In polar form t 0<= theta<= pi. 0<= r < = 2. Is it true? $\endgroup$ – Dore May 16 at 17:45
  • $\begingroup$ In polar, angle theta varies between $0$ and $2\pi$ $\endgroup$ – popi May 16 at 21:50
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Using divergence theorem:

$$\int_0^{2\pi}\int_0^2\int_{r\cos\theta}^8 2(1+r\sin \theta)\,dz\,dr\,d\theta=\boxed{64\pi}$$

By definition:

1.- Over plane $S_1\equiv z=x$: $\hspace{0.5cm} \displaystyle \int\int_{D_1}(x,y^2,x+y)(1,0,-1)dx\,dy=-\int\int_{D_1}y \,dx\,dy=\cdots=\boxed{0}$

with $D_1$ the circle $x^2+y^2\leq 4$.

2.- Over cylinder surface $S_2\equiv z=\pm\sqrt{4-x^2}$:$$ \int\int_{D_2}(x,4-x^2,z+\sqrt{4-x^2})\left(\frac{x}{\sqrt{4-x^2}},1,0\right)dx\,dz\hspace{0.3cm}+$$

$$+\hspace{0.3cm}\int\int_{D_2}(x,4-x^2,z-\sqrt{4-x^2})\left(\frac{x}{\sqrt{4-x^2}},-1,0\right)dx\,dz=\int\int_{D_2}\frac{2x^2}{\sqrt{4-x^2}}dz\,dx=$$

$$\int_{-2}^2\int_x^8\frac{2x^2}{\sqrt{4-x^2}}dz\,dx=\cdots=\boxed{32\pi}$$

3.- Over plane $S_3\equiv z=8$: $\hspace{0.5cm} \displaystyle \int\int_{D_1}(x,y^2,8+y)(0,0,1)dx\,dy=\int\int_{D_1}8+y \,dx\,dy=\int_0^{2\pi}\int_0^28+r\sin\theta\,dr\,d\theta=\boxed{32\pi}$

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