# Verify Divergence Theorem for bounded cylinder

$$F=xi+y^2j+(z+y)k$$ 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.

• What did you try? – DiegoMath May 16 at 17:25
• 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? – Dore May 16 at 17:45
• In polar, angle theta varies between $0$ and $2\pi$ – popi May 16 at 21:50

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