Question

(1) Let P denote the solid bounded by the surface of the hemisphere zV1--y2 and the cone z-Vx2 + y2 and let n denote an outwardly directed unit normal vector. Define the vector field (a) Evaluate the surface integral F nds directly without using Gauss' Divergence T heorem (b) Evaluate thetriplengral IIdiv(F) dV directly without using Gauss Diver- gence Theorem. confirming the result of Gauss' Divergence Theorem for this particular example.

Answer 1

Solution:

a)

outward normal for the sphere is

rl 2

r-cos θ r-sin θ _-.,T In cylindrical coordinates, n

2π FluxsphereJo

2T lid.A FluxsphereJo ' 0

2π FluxsphereJo Jo

r.2 cos θ-, r-sin_.r<) drd@ FluxsphereJo

2π r 〉 drd@ FluxsphereJo Jo

$\text{Flux}_{\text{sphere}}=\int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\left ( \frac{r^3\sin \theta\cos \theta}{\sqrt{1-r^2}}+r^3\sin^2 \theta+2r^2\cos \theta \right ) \mathrm{d}r \mathrm{d}\theta$

de 2π cos (0) sin (0) Fluxsphere

·2 cos FluXplereH3sin2(0) + (32-5-21)cos(9)sin(0) + 2;cos(2) Fluxsphere 48

3 sin (20)-24 sin (0) + (64-5-2) cos2 (0)-6θ [ 1 Fluxsp here - 192

$\text{Flux}_{\text{sphere}}=\frac{\pi}{16}$

$\text{outward normal for the cone is, }$

$\text{In cylindrical coordinates, }\vec{n}=\left \langle {-r\cos \theta},{-r\sin \theta},r \right \rangle$

$\text{Flux}_{\text{cone}}=\int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\vec{\text{F}}\cdot \hat{n}\mathrm{d}S$

$\text{Flux}_{\text{cone}}=\int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\vec{\text{F}}\cdot \frac{\vec{n}}{\left \| \vec{n} \right \|}\left \| \vec{n} \right \|\mathrm{d}A$

$\text{Flux}_{\text{cone}}=\int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\vec{\text{F}}\cdot \vec{n}\mathrm{d}A$

$\text{Flux}_{\text{cone}}=\int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\vec{\text{F}}\cdot \left \langle -r\cos \theta, -r\sin \theta, r \right \rangle \mathrm{d}r \mathrm{d}\theta$

$\text{Flux}_{\text{cone}}=\int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2}}}\left \langle r\sin \theta,r\sqrt{1-r^2}\sin \theta,2r\cos \theta \right \rangle\cdot \left \langle -r\cos \theta, -r\sin \theta, r \right \rangle \mathrm{d}r \mathrm{d}\theta$

2n 2V1-r2 sin2 θ + 2,2 cos θ) drdθ Fluxcone-

$\text{Flux}_{\text{cone}}=\int_0^{2\pi}\left [ -\sin^2\left({\theta}\right)\left(\dfrac{\arcsin\left(r\right)}{8}-\dfrac{r\left(1-r^2\right)^\frac{3}{2}}{4}+\dfrac{r\sqrt{1-r^2}}{8}\right)-\dfrac{\cos\left({\theta}\right)\sin\left({\theta}\right)r^3}{3}+\dfrac{2\cos\left({\theta}\right)r^3}{3} \right ]_0^{\frac{1}{\sqrt{2}}} \mathrm{d}\theta$

$\text{Flux}_{\text{cone}}=\int_0^{2\pi}\left ( -\dfrac{3{\pi}\sin^2\left({\theta}\right)+2^\frac{7}{2}\cos\left({\theta}\right)\sin\left({\theta}\right)-2^\frac{9}{2}\cos\left({\theta}\right)}{96} \right )\mathrm{d}\theta$

$\text{Flux}_{\text{cone}}=\left [ \dfrac{3{\pi}\sin\left(2{\theta}\right)+2^\frac{13}{2}\sin\left({\theta}\right)+2^\frac{9}{2}\cos^2\left({\theta}\right)-6{\pi}{\theta}}{384} \right ]_0^{2\pi}$

$\text{Flux}_{\text{cone}}=\frac{\pi}{16}$

$\text{Therefore, Flux}_{\text{total}}=\text{Flux}_{\text{sphere}}+\text{Flux}_{\text{cone}}=\frac{\pi}{16}+\frac{\pi}{16}=\frac{\pi}{8}$

$\text{b)}$

$\text{Flux}_{\text{total}}=\int \int \int_E\text{div}\vec{\text{F}}\mathrm{d}V$

Fluxtotal-

$\text{Flux}_{\text{total}}=\int \int \int_Ez\mathrm{d}V$

$\text{Flux}_{\text{total}}=\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_0^1\rho \cos \phi \left ( \rho^2 \sin \phi \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta \right )$

$\text{Flux}_{\text{total}}=\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_0^1\rho^3 \sin \phi \cos \phi \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta$

$\text{Flux}_{\text{total}}=\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \left [ \frac{\rho^4}{4} \right ]_0^1 \sin \phi \cos \phi \mathrm{d}\phi \mathrm{d}\theta$

$\text{Flux}_{\text{total}}=\frac{1}{4}\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \sin \phi \cos \phi \mathrm{d}\phi \mathrm{d}\theta$

$\text{Flux}_{\text{total}}=\frac{1}{4}\int_0^{2\pi}\left [ -\dfrac{\cos^2\left({\varphi}\right)}{2} \right ]_0^{\frac{\pi}{4}} \mathrm{d}\theta$

$\text{Flux}_{\text{total}}=\frac{1}{16}\int_0^{2\pi} \mathrm{d}\theta$

$\text{Flux}_{\text{total}}=\frac{2\pi}{16}=\frac{\pi}{8}$