Tools≻Tasks≻Browse:
Calculus - Vector≻Integration≻Flux≻3-D≻Through a Surface Defined over a Disk

Table 9.6.14(a)   Solution by task template

Initialize

Loading Student:-MultivariateCalculus

Define the vector field F

$⟨x\,y\,z⟩$$\stackrel{\text{assign to a name}}{\to }$$F$

Define the surface

$x^2\+y^2$$\stackrel{\text{assign to a name}}{\to }$$f$

Obtain N, a unit normal on the surface

$⟨-\frac{\partial }{\partial x} f\,-\frac{\partial }{\partial y} f\,1⟩$ = $\stackrel{\text{normalize}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$N$

Obtain $\mathrm{dsig}\=\sqrt{1\+f_x^2\+f_y^2}$ 

$\sqrt{1\+{\left(\frac{\partial }{\partial x} f\right)}^2\+{\left(\frac{\partial }{\partial y} f\right)}^2}$ = $\sqrt{4x^2\+4y^2\+1}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{dsig}$$$

Form $\mathbf{F}·\mathbf{N} \mathrm{dsig}$ and evaluate it on the surface

$\genfrac{}{}{0ex}{}{\mathbf{F}·\mathbf{N} \mathrm{dsig}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{z\=f}$

$\left(-\frac{2x^2}{\sqrt{4x^2\+4y^2\+1}}-\frac{2y^2}{\sqrt{4x^2\+4y^2\+1}}\+\frac{x^2\+y^2}{\sqrt{4x^2\+4y^2\+1}}\right)\sqrt{4x^2\+4y^2\+1}$

$\stackrel{\text{simplify}}{\=}$

$-x^2-y^2$

Define a change of coordinates and obtain its Jacobian

$1\+r \cos \left(\mathrm{\θ}\right)$$\stackrel{\text{assign to a name}}{\to }$$X$

$2\+r \sin \left(\mathrm{\θ}\right)$$\stackrel{\text{assign to a name}}{\to }$$Y$

$X\,Y$ = $1\+r\cos \left(\mathrm{\θ}\right)\,2\+r\sin \left(\mathrm{\θ}\right)$$\stackrel{\text{Jacobian}}{\to }$ $\stackrel{\text{determinant}}{\to }$${\cos \left(\mathrm{\θ}\right)}^{2}r\+r{\sin \left(\mathrm{\θ}\right)}^2$$\stackrel{\text{simplify}}{\=}$$r$$$

Form and evaluate the flux integral

$-{\int }_0^3{\int }_0^{2 \mathrm{\π}}r \left(X^2\+Y^2\right) \mathit{\ⅆ}\mathrm{\θ} \mathit{\ⅆ}r$ = $-\frac{171}{2}\mathrm{\π}$$$$$

Table 9.6.14(b)   Solution from first principles

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\:$

$\mathbf{F}≔\mathrm{VectorField}\left(⟨x\,y\,z⟩\right)\:$

Form and evaluate the flux integral with the Flux command

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Surface}\left(⟨x\,y\,x^2\+y^2⟩\,\left[x\,y\right]\=\mathrm{Circle}\left(⟨1\,2⟩\,3\,\left[r\,\mathrm{\θ}\right]\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^3{\∫}_0^{2\mathrm{\π}}r\left(-r^2-2r\cos \left(\mathrm{\θ}\right)-4r\sin \left(\mathrm{\θ}\right)-5\right)\ⅆ\mathrm{\θ}\ⅆr$

$\mathrm{Flux}\left(\mathbf{F}\,\mathrm{Surface}\left(⟨x\,y\,x^2\+y^2⟩\,\left[x\,y\right]\=\mathrm{Circle}\left(⟨1\,2⟩\,3\,\left[r\,\mathrm{\theta }\right]\right)\right)\right)$

$-\frac{171}{2}\mathrm{\π}$

Table 9.6.14(c)   Solution via the Flux command

Initialize

$\mathrm{BasisFormat}\left(\mathrm{false}\right)\:$

$f≔x^2\+y^2\:$

Obtain N, a unit normal on the surface

$\mathbf{N}≔\mathrm{Normalize}\left(⟨-\mathrm{diff}\left(f\,x\right)\,-\mathrm{diff}\left(f\,y\right)\,1⟩\right)$

Obtain $\mathrm{dsig}\=\sqrt{1\+f_x^2\+f_y^2}$ 

$\mathrm{dsig}≔\sqrt{1\+\mathrm{diff}{\left(f\,x\right)}^2\+\mathrm{diff}{\left(f\,y\right)}^2}$

$\sqrt{4x^2\+4y^2\+1}$

Obtain $\mathbf{F}·\mathbf{N} \mathrm{dsig}$ and evaluate it on the surface

$q≔\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{DotProduct}\left(\mathbf{F}\,\mathbf{N}\right) \mathrm{dsig}\,z\=f\right)\right)$ = $-x^2-y^2$$$$$

Change coordinates

$Q≔\mathrm{eval}\left(q\,\left[x\=1\+r \cos \left(\mathrm{\θ}\right)\,y\=2\+r \sin \left(\mathrm{\θ}\right)\right]\right)$ = $-{\left(1\+r\cos \left(\mathrm{\θ}\right)\right)}^2-{\left(2\+r\sin \left(\mathrm{\θ}\right)\right)}^2$$$

$\mathrm{simplify}\left(\mathrm{Jacobian}\left(\left[1\+r \cos \left(\mathrm{\theta }\right)\,2\+r \sin \left(\mathrm{\theta }\right)\right]\,\left[r\,\mathrm{\θ}\right]\,\mathrm{determinant}\=\mathrm{true}\right)\left[2\right]\right)$ = $r$$$

Use the Int command to form the flux integral and int to evaluate it.

$\mathrm{Int}\left(r Q\,\left[ \mathrm{\θ}\=0..2 \mathrm{\π}\,r\=0..3\right]\right)\= :-\mathrm{int}\left(r Q\,\left[\mathrm{\θ}\=0..2 \mathrm{\π}\,r\=0..3\right]\right)$

${\∫}_0^3{\∫}_0^{2\mathrm{\π}}r\left(-{\left(1\+r\cos \left(\mathrm{\θ}\right)\right)}^2-{\left(2\+r\sin \left(\mathrm{\θ}\right)\right)}^2\right)\ⅆ\mathrm{\θ}\ⅆr\=-\frac{171}{2}\mathrm{\π}$

Table 9.6.14(d)   Solution from first principles