Question

Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+2z)j + (3z)k be a vector field (for example, the velocityfaild of a fluid flow). the solid object has five sides, S1:bottom(xy-plane), S2:left side(xz-plane), S3 rear side(yz-plane), S4:right side, and S5:cylindrical roof.

a. Sketch the solid object.
b. Evaluate the flux of F through each side of the object (S1,S2,S3,S4,S5).
c. Find the total flux through surface S.

Answer 1

a. I've attached a plot of the surface. Each face is parameterized by

•
`\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j` with
`0\le x\le2` and
`0\le y\le6-x`;

•
`\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k` with
`0\le u\le2` and
`0\le v\le\frac\pi2`;

•
`\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k` with
`0\le y\le 6` and
`0\le z\le2`;

•
`\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k` with
`0\le u\le2` and
`0\le v\le\frac\pi2`; and

•
`\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k` with
`0\le u\le\frac\pi2` and
`0\le y\le6-2\cos u`.

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

`\mathbf n_1=(\partial\mathbf s_1)/(\partial y)*(\partial\mathbf s_1)/(\partial x)=-\mathbf k`

`\mathbf n_2=(\partial\mathbf s_2)/(\partial u)*(\partial\mathbf s_2)/(\partial v)=-u\,\mathbf j`

`\mathbf n_3=(\partial\mathbf s_3)/(\partial z)*(\partial\mathbf s_3)/(\partial y)=-\mathbf i`

`\mathbf n_4=(\partial\mathbf s_4)/(\partial v)*(\partial\mathbf s_4)/(\partial u)=u\,\mathbf i+u\,\mathbf j`

`\mathbf n_5=(\partial\mathbf s_5)/(\partial y)*(\partial\mathbf s_5)/(\partial u)=2\cos u\,\mathbf i+2\sin u\,\mathbf k`

Then integrate the dot product of f with each normal vector over the corresponding face.

`\displaystyle\iint_(S_1)\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^(6-x)f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx`

`=\displaystyle\int_0^2\int_0^(6-x)0\,\mathrm dy\,\mathrm dx=0`

`\displaystyle\iint_(S_2)\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^(\frac\pi2)\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du`

`\displaystyle=\int_0^2\int_0^(\frac\pi2)-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8`

`\displaystyle\iint_(S_3)\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz`

`=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0`

`\displaystyle\iint_(S_4)\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^(\frac\pi2)\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du`

`=\displaystyle\int_0^2\int_0^(\frac\pi2)-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi`

`\displaystyle\iint_(S_5)\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^(\frac\pi2)\int_0^(6-2\cos u)\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du`

`=\displaystyle\int_0^(\frac\pi2)\int_0^(6-2\cos u)12\,\mathrm dy\,\mathrm du=36\pi-24`

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since S is closed, we can find the total flux by applying the divergence theorem.

`\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV`

where R is the interior of S. We have

`\mathrm{div}\mathbf f(x,y,z)=(\partial(3x))/(\partial x)+(\partial(x+y+2z))/(\partial y)+(\partial(3z))/(\partial z)=7`

The integral is easily computed in cylindrical coordinates:

`\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\frac\pi2`

`\displaystyle\int_0^2\int_0^(\frac\pi2)\int_0^(6-r\cos t)7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3`

as expected.