OBJECTIVES

When you finish this section, you should be able to:

1. Find a surface integral using a double integral (p. 1027)
2. Determine the orientation of a surface (p. 1030)
3. Find the flux of a vector field across a surface (p. 1033)
   Application: Electric Flux

spanDEFINITIONspan Surface Integral

Let \(S\) be a surface defined on a closed, bounded region \(R\) with a smooth parametrization \(\mathbf{r}(u,v) =x(u,v) \mathbf{i}+y(u,v) \mathbf{j} +z(u,v) \mathbf{k}\). Also, let \(w= F(x,y,z)\) be defined over a solid containing the surface \(S.\) Then the surface integral of \(F\) over \(S\) is defined as \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \iint\limits_{\kern-3ptS}F(x,y,z)\,dS=\lim\limits_{\Vert \Delta \Vert \rightarrow 0}\sum\limits_{k=1}^{n} F\left(x\left( u_{k}^{\ast },v_{k}^{\ast }\right)\, y\left(u_{k}^{\ast },v_{k}^{\ast }\right)\,z\left( u_{k}^{\ast },v_{k}^{\ast }\right) \right) \Delta S_{k} }} \]

provided this limit exists.

THEOREM A Surface Integral Expressed as a Double Integral

Let \(S\) be a surface defined on a closed, bounded region \(R\) with a smooth parametrization \(\mathbf{r}(u,v) =x(u,v) \mathbf{i}+y(u,v) \mathbf{j} +z(u,v) \mathbf{k}\). Also, suppose that \(w=F(x,y,z)\) is continuous on a solid containing the surface \(S\). Then the surface integral of \(F\) over \(S\) is given by \[ \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\iint\limits_{\kern-3ptS}F(x,y,z)\,dS=\iint\limits_{\kern-3ptR}F(x(u,v) ,y(u,v) ,\,z(u,v))\left\Vert \mathbf{r} _{u}\times \mathbf{r}_{v}\right\Vert \,du\,dv } } \tag{1} \end{equation*} \]

Finding a Surface Integral

Find \(\iint\limits_{\kern-3ptS}ye^{x^{2}+z^{2}}dS,\) where \(S\) is the part of the surface of the cylinder \(x^{2}+z^{2}=9\) that lies between \(y=0\) and \(y=2.\)

Solution The surface \(S\) is shown in Figure 52. Since the surface is a piece of a cylinder, we use a variation of cylindrical coordinates to parametrize the surface. Here, we use \(x=r\cos \theta ,\) \(y=y,\) and \(z=r\sin \theta .\) (Do you see why?) Then \(r^{2}=x^{2}+z^{2}=9,\) so \(r=3.\) We parametrize the surface \(S\) using \[ \begin{equation*} \mathbf{r}( \theta ,y) =3\cos \theta\ \mathbf{i}+y\mathbf{j}+3\sin \theta\ \mathbf{k} \end{equation*} \]

where the parameter domain \({R}\) is \(0\leq \theta \leq 2\pi \) and \(0\leq y\leq 2.\)

The partial derivatives of \(\mathbf{r}\) are \[ \mathbf{r}_{\theta }( \theta ,y) =-3\sin \theta\ \mathbf{i}+3\cos \theta\ \mathbf{k}\qquad\hbox{and}\qquad \mathbf{r}_{y}( \theta ,y) =\mathbf{j} \]

and their cross product is \[ \begin{equation*} \mathbf{r}_{\theta }\times \mathbf{r}_{y}= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3\sin \theta & 0 & 3\cos \theta \\ 0 & 1 & 0 \end{array}\right| =-3\cos \theta\ \mathbf{i}-3\sin \theta\ \mathbf{k} \end{equation*} \]

Then \(\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{y}\right\Vert =\sqrt{9\cos ^{2}\theta +9\sin ^{2}\theta }=3.\) So, we have \[ \begin{eqnarray*} \iint\limits_{\kern-3ptS}ye^{x^{2}+z^{2}}dS &=&\iint\limits_{\kern-3ptR}ye^{9}\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{y}\right\Vert d\theta dy=\int_{0}^{2}\int_{0}^{2\pi }( ye^{9}) 3\,d\theta\, dy \notag \\[4pt] &=&3e^{9}\int_{0}^{2}y\big[ \theta \big] _{0}^{2\pi }dy=3e^{9}\int_{0}^{2}2\pi y\,dy=6\pi e^{9}\left[ \dfrac{y^{2}}{2}\right] _{0}^{2}=12\pi e^{9} \end{eqnarray*} \]

RECALL

A lamina is a thin, flat sheet of material.

NOW WORK

Problem 3.

NEED TO REVIEW?

The center of mass of a lamina with variable density is discussed in Section 14.4, pp. 929-932.

Finding the Mass of a Lamina

A lamina in the shape of the cone \(z=6-\sqrt{x^{2}+y^{2}}\) lies between the planes \(z=2\) and \(z=5\). If the mass density of the lamina is \(\rho (x,y,z)=\sqrt{x^{2}+y^{2}}\), find the mass \(M\) of the lamina.

Figure 53

Solution Figure 53 illustrates the lamina. Using \(x=r\cos \theta , \) \(y=r\sin \theta ,\) and \(z=6-\sqrt{x^{2}+y^{2}}=6-r,\) we parametrize the surface as \[ \mathbf{r}=\mathbf{r}\left( r,\theta \right) =r\cos \theta \,\mathbf{i} +r\sin \theta \,\mathbf{j}+\left( 6-r\right) \mathbf{k}\qquad 0\leq \theta \leq 2\pi\quad 1\leq r\leq 4 \]

First we find \(\mathbf{r}_{r}\times \mathbf{r}_{\theta }.\) \[ \begin{eqnarray*} \mathbf{r}_{r}\times \mathbf{r}_{\theta } &=& \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos \theta & \sin \theta & -1 \\ -r\sin \theta & r\cos \theta & 0 \end{array}\right| =r\cos \theta \,\mathbf{i} + r\sin \theta \,\mathbf{j}+\left( r\cos ^{2}\theta +r\sin ^{2}\theta \right) \mathbf{k}\\ &=&r\cos \theta \,\mathbf{i}+r\sin \theta \,\mathbf{j}+r\,\mathbf{k} \\[4pt] \hspace{-2pt}\left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert &=&\sqrt{r^{2}\cos ^{2}\theta +r^{2}\sin ^{2}\theta +r^{2}}=r\sqrt{\cos ^{2}\theta +\sin ^{2}\theta +1}=r\sqrt{1+1}=\sqrt{2}\,r \end{eqnarray*} \]

Since the mass density of the lamina is \(\rho (x,y,z)=\sqrt{x^{2}+y^{2}}=r\) and \(dS=\left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert dr\,d\theta =\sqrt{2}r\,dr\,d\theta ,\) the mass \(M\) of the lamina is given by \[ \begin{eqnarray*} M &=&\iint\limits_{\kern-3ptS}\rho (x,y,z)\,dS=\iint\limits_{\kern-3ptR}\rho \left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert dr\,d\theta =\iint\limits_{\kern-3ptR}r(\sqrt{2}r) dr\,d\theta \\ &=&\sqrt{2}\int_{0}^{2\pi }\int_{1}^{4}r^{2}\,dr\,d\theta =\sqrt{2} \int_{0}^{2\pi }\left[ \dfrac{r^{3}}{3}\right] _{1}^{4}\,d\theta =21\sqrt{2} \int_{0}^{2\pi }d\theta =42\sqrt{2}\,\pi \end{eqnarray*} \]

NOW WORK

Problem 33.

RECALL

The centroid is the geometric center of a lamina, and it is numerically equivalent to the center of mass of a homogeneous lamina that has mass density \(\rho =1.\)

Finding the Centroid of a Surface

Find the centroid of the part of the unit sphere that lies in the first octant. See Figure 54.

Figure 54

Solution In spherical coordinates, the equation of the unit sphere \(S\): \(x^{2}+y^{2}+z^{2}=1\) is \(\rho =1.\) So using \(x=\cos \theta \sin \phi \), \(y=\sin \theta \sin \phi \), and \(z=\cos \phi\), we obtain the parametrization \[ \mathbf{r}=\mathbf{r}\left( \theta ,\phi \right) =\cos \theta \sin \phi \, \mathbf{i}+\sin \theta \sin \phi \,\mathbf{j}+\cos \phi\, \mathbf{k};\qquad 0\leq \theta \leq \dfrac{\pi }{2},\quad 0\leq \phi \leq \dfrac{\pi }{2} \]

We seek \(\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{\phi }\right\Vert \) and \(dS.\) The partial derivatives of \(\mathbf{r}=\mathbf{r}\left( \theta ,\phi \right) \) are \[ \begin{eqnarray*} \mathbf{r}_{\theta }&=&-\sin \theta \sin \phi \,\mathbf{i}+\cos \theta \sin \phi \,\mathbf{j} \,\quad \mathbf{r}_{\phi }=\cos \theta \cos \phi \,\mathbf{i}+\sin \theta \cos \phi \,\mathbf{j}-\sin \phi\, \mathbf{k}\\[4pt] \mathbf{r}_{\theta }\times \mathbf{r}_{\phi } &=&\left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -\sin \theta \sin \phi & \cos \theta \sin \phi & 0 \\[3pt] \cos \theta \cos \phi & \sin \theta \cos \phi & -\sin \phi \end{array}\right|\\[4pt] &=&-\cos \theta \sin ^{2}\phi \,\mathbf{i}-\sin \theta \sin ^{2}\phi \,\mathbf{ j}-\sin \phi \cos \phi\, \mathbf{k} \\[4pt] \left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{\phi }\right\Vert &=&\sqrt{ \cos ^{2}\theta \sin ^{4}\phi +\sin ^{2}\theta \sin ^{4}\phi +\sin ^{2}\phi \cos ^{2}\phi }\\[4pt] &=&\sqrt{( \cos ^{2}\theta +\sin ^{2}\theta ) \sin ^{4}\phi +\sin ^{2}\phi \cos ^{2}\phi } \\[4pt] &=&\sqrt{\sin ^{2}\phi (\sin ^{2}\phi +\cos ^{2}\phi) }=\sin \phi\\[4pt] dS&=&\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{\phi }\right\Vert d\theta\, d\phi =\sin \phi \,d\theta \,d\phi \end{eqnarray*} \]

Since the surface area \(S\) of a sphere is \(S=4\pi r^{2},\) the surface area of the part of the unit sphere \(r=1\) in the first octant is \(\dfrac{1}{8}S=\dfrac{\pi }{2},\) so the mass \(M\) is \(\dfrac{\pi }{2}.\) Because of symmetry, \(\overline{x}=\overline{y}=\overline{z},\) so we need to find only one moment, say \(M_{xy}.\) \[ \begin{eqnarray*} M_{xy}&=&\iint\limits_{\kern-3ptS}z\,dS=\iint\limits_{\kern-3ptS}\cos \phi \,dS=\int_{0}^{\pi /2}\int_{0}^{\pi /2}\cos \phi\, \sin \phi\, d\theta\, d\phi\\[4pt] &=&\int_{0}^{\pi /2}\dfrac{\pi }{2}\cos \phi\, \sin \phi\, d\phi =\dfrac{\pi }{2}\left[ \dfrac{\sin ^{2}\phi }{2}\right] _{0}^{\pi /2}=\dfrac{\pi }{4} \end{eqnarray*} \]

The centroid of the part of the unit sphere in the first octant is \[ \left( \overline{x},\overline{y},\overline{z}\right) =\left( \dfrac{M_{yz}}{M },\dfrac{M_{xz}}{M},\dfrac{M_{xy}}{M}\right) =\left( \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2} \right) \]

NEED TO REVIEW?

Unit normal vectors are discussed in Section 11.2, pp. 769-770.

NOTE

When \(F( x,y,z) =z-f(x,y)\), then \(S\) is the level surface \(F( x,y,z) =0\). Level surfaces are discussed in Section 12.1, pp. 812-815.

Finding the Orientations of a Surface

Find both orientations for the paraboloid defined by \(z=9-x^{2}-y^{2},\) \(z\geq 0.\)

1. Use rectangular coordinates to find the unit normal vectors.
2. Parametrize the surface, and then find the unit normal vectors.

Solution (a) Since the paraboloid is given by \(f(x,y) = 9-x^{2} - y^{2}, z \ge 0\), we can use the equations (3) and (4) to find the unit normal vectors. We begin by finding the partial derivatives \[ f_{x}(x,y) =-2x \qquad f_{y}(x,y) =-2y \]

Then \[ \begin{eqnarray*} \mathbf{n}&=&\dfrac{2x\,\mathbf{i}+2y\,\mathbf{j}+\mathbf{k}}{\sqrt{[-2x]^{2}+[-2y]^{2}+1}}=\dfrac{2x\,\mathbf{i}+2y\,\mathbf{j}+\mathbf{k}}{\sqrt{4x^{2}+4y^{2}+1}} \\[7pt] -\mathbf{n}&=&\dfrac{-2x\,\mathbf{i-}2y\,\mathbf{j}-\mathbf{k}}{\sqrt{[-2x]^{2}+[-2y]^{2}+1}}=\dfrac{ -2x\,\mathbf{i}-2y\,\mathbf{j}-\mathbf{k}}{\sqrt{4x^{2}+4y^{2}+1}} \end{eqnarray*} \]

Notice that the \(\mathbf{k}\) component of \(\mathbf{n}\) is positive, indicating \(\mathbf{n}\) is the upward-pointing unit normal vector.

(b) We parametrize \(z=9-x^{2}-y^{2},\) \(z\geq 0\), using cylindrical coordinates, by defining the parametric equations \[ \begin{equation*} x=r\cos \theta \qquad y=r\sin \theta \qquad z=9-(x^{2}+y^{2}) =9-r^{2} \end{equation*} \]

Then the parametrization is \[ \begin{equation*} \mathbf{r}=\mathbf{r}( r,\theta ) =r\cos \theta \,\mathbf{i} +r\sin \theta \,\mathbf{j}+( 9-r^{2}) \mathbf{k} \end{equation*} \]

We seek \(\left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert \). The partial derivatives of \(\mathbf{r}\) are \[ \begin{eqnarray*} \mathbf{r}_{r}&=&\cos \theta \,\mathbf{i}+\sin \theta \,\mathbf{j}-2r\mathbf{k} \qquad \mathbf{r}_{\theta }=-r\sin \theta \,\mathbf{i}+r\cos \theta \,\mathbf{j}\\[4pt] \mathbf{r}_{r}\times \mathbf{r}_{\theta } &=& \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \cos \theta & \sin \theta & -2r \\[3pt] -r\sin \theta & r\cos \theta & 0 \end{array}\right| =2r^{2}\cos \theta\, \mathbf{i}+2r^{2}\sin \theta\, \mathbf{j}+r\mathbf{k} \\[4pt] \left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert &=&\sqrt{ 4r^{4}\cos ^{2}\theta +4r^{4}\sin ^{2}\theta +r^{2}}=\sqrt{4r^{4}+r^{2}}=r \sqrt{4r^{2}+1} \end{eqnarray*} \]

The unit normal vectors are \[ \begin{eqnarray*} \mathbf{n}&=&\dfrac{\mathbf{r}_{r}\times \mathbf{r}_{\theta }}{\left\Vert \mathbf{r}_{r}\times \mathbf{r}_{\theta }\right\Vert }=\dfrac{2r^{2}\cos \theta\, \mathbf{i}+2r^{2}\sin \theta\, \mathbf{j}+r\mathbf{k}}{r\sqrt{4r^{2}+1}} =\dfrac{2r\cos \theta\, \mathbf{i}+2r\sin \theta\, \mathbf{j}+\mathbf{k}}{\sqrt{ 4r^{2}+1}}\\[4pt] -\mathbf{n}&=&\dfrac{-2r\cos \theta\, \mathbf{i} -2r\sin \theta\, \mathbf{j}-\mathbf{k}}{\sqrt{4r^{2}+1}} \end{eqnarray*} \]

Figure 58 The outer unit normal vectors of \(z=9-x^{2}-y^{2},\) \(z\geq 0.\)

Notice that the unit vectors found in (a) and (b) are equivalent.

NOW WORK

Problem 17.

NOTE

A Möbius strip is formed by taking a long rectangular strip of paper, giving one end a half-twist, and attaching the two ends. The Möbius strip is named after the German mathematician August Möbius (1790–1868).

Finding the \(\mathbf{k}\) Component of the Outer Unit Normal Vectors of an \(xy\,\)-Simple Surface

Consider the surface \(S\) with positive orientation that forms the boundary of the closed, bounded solid \(E\) shown in Figure 61. Notice that the surface \(S\) can be decomposed into three surfaces, \(S_{1},\) \(S_{2},\) and \(S_{3}\), and that each of the outer unit normal vectors \(\mathbf{n}_{1},\) \(\mathbf{n}_{2}\), and \(\mathbf{n}_{3}\) of the three surfaces points away from the solid \(E\) enclosed by \(S\).

Figure 61

Assume the surfaces \(S_{1}\) and \(S_{2}\) are defined by the equations \[ \begin{equation*} S_{1}{:}\quad z=f_{1}(x,y)\qquad \hbox{and}\qquad S_{2}{:}\quad z=f_{2}(x,y) \end{equation*} \]

where \(f_{1}(x,y)<f_{2}(x,y)\) on \(D,\) the functions \(f_{1}\) and \(f_{2}\) are continuous on \(D\), and both \(f_{1}\) and \(\ f_{2}\) have continuous partial derivatives on \(D\). The surface \(S_{3}\) is the portion of a cylindrical surface between \(S_{1}\) and \(S_{2}\) formed by lines parallel to the \(z\)-axis and along the boundary of \(D\). The solid \(E\) is enclosed by \(S_{2}\) on the top, \(S_{1}\) on the bottom, and the lateral surface \(S_{3}\) on the sides. A surface \(S\) with these properties is called \(\boldsymbol x\)\(\boldsymbol y\)-simple.

Find the \(\mathbf{k}\) component of the outer unit normal vectors of \(S\).

Solution For the top surface \(S_{2}\): \(z=f_{2}(x,y)\), the outer unit normal vector \(\mathbf{n}_{2}\) is the same as the upward-pointing unit normal vector to \(S_{2}\). \[ \begin{equation*} \mathbf{n}_{2}=\dfrac{-(f_{2})_x(x,y)\mathbf{i}-(f_{2})_ y(x,y)\mathbf{j}+\mathbf{k} }{\sqrt{[(f_{2})_x(x,y)]^{2}+[(f_{2})_y(x,y)]^{2}+1}} \end{equation*} \]

The \(\mathbf{k}\) component of \(\mathbf{n}_{2}\) is given by the direction cosine \(\cos \gamma _{2}\), where \[ \cos \gamma _{2}=\mathbf{n}_{2}\,{\cdot}\, \mathbf{k}=\dfrac{1}{\sqrt{ [(f_{2})_x(x,y)]^{2}+[(f_{2})_y(x,y)]^{2}+1}} \]

For the bottom surface \(S_{1}\): \(z=f_{1}(x,y)\), the outer unit normal vector \(\mathbf{n}_{1}\) equals the downward-pointing unit normal vector \(-\mathbf{n}\) to \(S_{1}\). \[ \begin{equation*} \mathbf{n}_{1}=-\mathbf{n}=\dfrac{(f_{1})_{x}(x,y)\mathbf{i}+(f_{1})_y(x,y)\mathbf{j}- \mathbf{k}}{\sqrt{[(f_{1})_x(x,y)]^{2}+[(f_{1})_y(x,y)]^{2}+1}} \end{equation*} \]

The \(\mathbf{k}\) component of \(\mathbf{n}_{1}\) is given by the direction cosine \(\cos \gamma _{1}\), where \[ \cos \gamma _{1}=\mathbf{n}_{1}\,{\cdot}\, \mathbf{k}=\dfrac{-1}{\sqrt{[(f_{1})_x(x,y)]^{2}+[(f_{1})_y(x,y)]^{2}+1}} \]

For the lateral surface \(S_{3}\), the outer unit normal vector \(\mathbf{n} _{3} \) is orthogonal to \(\mathbf{k}\) at each point on \(S_{3}\). The \(\mathbf{k }\) component of \(\mathbf{n}_{3},\) given by the direction cosine, \(\cos \gamma _{3}\), is therefore \(0\).

Finding the Outer Unit Normal Vectors to a Surface \(S\)

Find the outer unit normal vectors to the solid \(E\) enclosed by \[ z=f(x,y)=\sqrt{R^{2}-x^{2}-y^{2}}\quad\hbox{and}\quad z=0\qquad 0\leq x^{2}+y^{2}\leq R^{2} \]

Figure 62

Solution The solid \(E\) is the interior of a hemisphere with center at \((0,0,0)\) and radius \(R\) as shown in Figure 62. The surface \(S\) consists of two surfaces, \(S_{1}\) and \(S_{2}\). The bottom surface \(S_{1}\) and top surface \(S_{2}\) are defined by \[ \begin{equation*} S_{1}{:}\quad z=0 \quad \hbox{and}\quad S_{2}{:}\quad z=f(x,y)=\sqrt{R^{2}-x^{2}-y^{2}}\qquad 0\leq x^{2}+y^{2}\leq R^{2} \end{equation*} \]

The outer unit normal vector \(\mathbf{n}_{1}\) of \(S_{1}\) is \(-\mathbf{k}\).

To find the outer unit normal vector \(\mathbf{n}_{2}\) of \(S_{2}\), we find \(f_{x}(x,y) \) and \(f_{y}(x,y) .\) \[ \begin{equation*} f_{x}(x,y)=\dfrac{-x}{\sqrt{R^{2}-x^{2}-y^{2}}}=\dfrac{-x}{z}\qquad \hbox{and} \qquad f_{y}(x,y)=\dfrac{-y}{\sqrt{R^{2}-x^{2}-y^{2}}}=\dfrac{-y}{z} \end{equation*} \]

Then \[ \begin{eqnarray*} \mathbf{n}_{2}&=&\dfrac{-f_{x}(x,y)\mathbf{i}-f_{y}(x,y)\mathbf{j}+\mathbf{k}}{ \sqrt{[f_{x}(x,y)]^{2}+[f_{y}(x,y)]^{2}+1}}=\dfrac{\dfrac{x}{z}\mathbf{i}+ \dfrac{y}{z}\mathbf{j}+\mathbf{k}}{\sqrt{\dfrac{x^{2}}{z^{2}}+\dfrac{y^{2}}{ z^{2}}+1}}\\ &=&\dfrac{x\mathbf{i}+y\mathbf{j}+z\,\mathbf{k}}{\sqrt{ x^{2}+y^{2}+z^{2}}}=\dfrac{x\mathbf{i}+y\mathbf{i}+z\mathbf{k}}{R} \end{eqnarray*} \]

NOW WORK

Problem 13.

Finding the Flux of \(\mathbf{F}\) Across a Cylinder

A fluid has a constant mass density \(\rho .\) Find the mass of fluid flowing across the cylinder \(x^{2}+y^{2}=4,\) where \(0\leq z\leq 3,\) in a unit of time in the direction outward from the \(z\)-axis, if the velocity of the fluid at any point on the cylinder is \(\mathbf{F}=\mathbf{F}(x,y,z)=x\mathbf{i} +y\mathbf{j}+2z\mathbf{k}\). That is, find the flux of \(\mathbf{F}\) across the cylinder.

Solution In cylindrical coordinates \(x^{2}+y^{2}=r^{2}=4,\) or equivalently, \(r=2,\) which leads to the parametrization \(\mathbf{r}\left( \theta ,z\right) =2\cos \theta \,\mathbf{i}+2\sin \theta \,\mathbf{j}+z\,\mathbf{k}\), \(0\leq \theta \leq 2\pi ,\) \(0\leq z\leq 3.\)

We seek \(\left\Vert \mathbf{r}_{\theta}\times \mathbf{r}_{z }\right\Vert \) and \(dS\).

The partial derivatives of \(\mathbf{r}\) are \[ \begin{eqnarray*} \mathbf{r}_{\theta }&=&-2\sin \theta \,\mathbf{i}+2\cos \theta \,\mathbf{j}\quad \quad \ \mathbf{r}_{z}=\mathbf{k}\\[4pt] \mathbf{r}_{\theta }\times \mathbf{r}_{z} &=& \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -2\sin \theta & 2\cos \theta & 0 \\[3pt] 0 & 0 & 1 \end{array}\right| =2\cos \theta\, \mathbf{i}+2\sin \theta\, \mathbf{j} \\[4pt] \left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{z}\right\Vert &=&\sqrt{ 4\cos ^{2}\theta +4\sin ^{2}\theta }=2 \\[4pt] dS &=&\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{z}\right\Vert d\theta \,dz=2\,d\theta \,dz \end{eqnarray*} \]

Figure 65 \(x^{2}+y^{2}=4,\) \(0\leq z\leq 3,\) and \(\mathbf{n}=\cos \theta\, \mathbf{i}+\sin \theta\, \mathbf{j}\)

Then the unit normal vector \(\mathbf{n}\) is \[ \begin{equation*} \mathbf{n}=\dfrac{\mathbf{r}_{\theta}\times \mathbf{r}_{z}}{\left\Vert \mathbf{r}_{\theta}\times \mathbf{r}_{z}\right\Vert }=\dfrac{2\cos \theta\, \mathbf{i}+2\sin \theta\, \mathbf{j}}{2}=\cos \theta\, \mathbf{i}+\sin \theta\, \mathbf{j} \end{equation*} \]

These vectors point outward from the surface of the cylinder, so we have the correct orientation. Figure 65 shows the cylinder \(x^{2}+y^{2}=4,\) where \(0\leq z\leq 3,\) and the outer unit normal vectors to the cylinder.

We now write \(\mathbf{F}\) using the parametrization. \[ \mathbf{F}=\mathbf{F}(x,y,z)=\mathbf{F}(\theta ,z)=2\cos \theta \,\mathbf{i} +2\sin \theta \,\mathbf{j}+2z\,\mathbf{k} \]

The velocity of the fluid in the direction of \(\mathbf{n}\) is \[ \begin{equation*} \mathbf{F\,{\cdot}\, n}=\left( 2\cos \theta \,\mathbf{i}+2\sin \theta \,\mathbf{j} +2z\,\mathbf{k}\right) \,{\cdot}\, \left( \cos \theta\, \mathbf{i}+\sin \theta\, \mathbf{j}\right) =2\cos ^{2}\theta \,+2\sin ^{2}\theta =2 \end{equation*} \]

The mass of fluid flowing across the cylinder \(x^{2}+y^{2}=4\) in a unit of time in the direction of \(\mathbf{n}\) is given by \[ \iint\limits_{\kern-3ptS}\rho \,\mathbf{F}\,{\cdot}\, \mathbf{n}\,dS=\int_{0}^{3} \int_{0}^{2\pi }\left( \rho \,2\right) 2\,d\theta \,dz=4\rho \int_{0}^{3}2\pi \,dz=\left( 8\rho \pi \right) 3=24\rho \pi \]

The flux of \(\mathbf{F}\) across \(S\) is \(24\rho \pi.\)

NOW WORK

Problem 39.

Finding the Flux of \(\mathbf{F}\) Across a Cube

A fluid has constant mass density \(\rho .\) Find the mass per unit time of fluid flowing across the cube enclosed by the planes \(x=0\), \(x=1\), \(y=0\), \(y=1\), \(z=0\), and \(z=1\) in the direction of the outer unit normal vectors if the velocity of the fluid at any point on the cube is \(\mathbf{F}=\mathbf{F} (x,y,z)=4xz\mathbf{i}-y^{2}\mathbf{j}+yz\mathbf{k}\). That is, find the flux of \(\mathbf{F}\) across the cube.

Figure 66

Solution Figure 66 shows the cube and its six faces. The mass per unit time of fluid flowing across the cube in the direction of the outer unit normal vector \(\mathbf{n}\) is \[ \iint\limits_{\kern-3ptS}\rho \mathbf{F}\,{\cdot}\, \mathbf{n}\,dS=\rho \iint\limits_{\kern-3ptS} \mathbf{F}\,{\cdot}\, \mathbf{n}\,dS \]

We decompose \(S\) into its six faces and find the surface integral over each one, as shown in Table 1.

The mass per unit time of fluid flowing across the cube is \[ 0+0+0+2\rho -\rho +\dfrac{1}{2}\rho = \dfrac{3}{2}\rho \]

NOW WORK

Problem 23.