Surface Integrals

The proof is omitted.

Let us first parametrize the surface. \begin{eqnarray*} x&=&3\cos\theta\sin\phi\\ y&=&3\sin\theta\sin\phi\\ z&=&3\cos\phi\\ 0\leq&\theta&\leq 2\pi\\ 0\leq&\phi&\leq \frac{\pi}2. \end{eqnarray*} Then we have \begin{eqnarray*}\overrightarrow R(\theta,\phi)&=&\langle 3\cos\theta\sin\phi,3 \sin\theta\sin\phi,3\cos\phi\rangle\\ \overrightarrow{R_{\theta}}(\theta,\phi)&=&\langle -3\sin\theta\sin\phi,3\cos\theta\sin\phi,0\rangle\\ \overrightarrow{R_{\phi}}(\theta,\phi)&=&\langle 3\cos\theta\cos\phi,3\sin\theta\cos\phi,-3\sin\phi\rangle\\ \overrightarrow{R_{\theta}}\times\overrightarrow{R_{\phi}}&=& \mbox{det }\left|\begin{array}{ccc} \overrightarrow i & \overrightarrow j &\overrightarrow k\newline -\sin\theta\sin\phi&\cos\theta\sin\phi&0\newline \cos\theta\cos\phi&\sin\theta\cos\phi&-\sin\phi \end{array}\right| \\ &=&-9\sin^2\phi\cos\theta\overrightarrow i-\sin^2\phi\sin\theta\overrightarrow j-\sin\phi\cos\phi\overrightarrow k\\ dS&=&\left|\overrightarrow{R_{\theta}}\times\overrightarrow{R_{\phi}}\right|\,d\theta d\phi \\ &=&9\sqrt{\sin^4\phi\cos^2\theta+\sin^4\phi\sin^2\theta+\sin^2\phi\cos^2\phi}\,d\theta d\phi \\ &=&9 |\sin\phi|\,d\theta d\phi\\ &=&9\sin\phi\,d\theta d\phi. \end{eqnarray*}

Thus, we have: \begin{eqnarray*} \iint_{\sigma}f\,dS&=&\iint_D( 3\cos\theta\sin\phi+3\sin\theta\sin\phi+3\cos\phi)\cdot 9 \sin\phi\,d\theta d\phi \\ &=&27\int_0^{\frac{\pi}2}\int_0^{2\pi} \left( \cos\theta\sin^2\phi+\sin\theta\sin^2\phi+\cos\phi \sin\phi\right)\,d\theta d\phi \\ &=&27\int_0^{\frac{\pi}2}\left.\left(\sin^2\phi\cdot \sin\theta\right)\right|_{\theta=0}^{2\pi}\,d\phi \\ &&- 27\int_0^{\frac{\pi}2}\left.\left(\sin^2\phi\cdot \cos\theta\right)\right|_{\theta=0}^{2\pi}\,d\phi \\ && + 2\pi\cdot 27\int_0^{\frac{\pi}2}\sin\phi\cos\phi\,d\phi\\ &=&-27\pi\cdot \left.\frac12\cos(2\phi)\right|_{0}^{\pi/2}\\ &=&27\pi. \end{eqnarray*}

We can parametrize the surface in the following way: \begin{eqnarray*} x&=&r\cos\theta\\ y&=&r\sin\theta\\ z&=&4-r^2\\ 0\leq&r&\leq 2\\ 0\leq&\theta&\leq 2\pi. \end{eqnarray*} After finding the parametrization we must find the normal vector \(\overrightarrow N\) that is consistent with the orientation. In order to do this we need \(\overrightarrow R\), \(\overrightarrow {R_{\theta}}\), and \(\overrightarrow{R_r}\). \begin{eqnarray*} \overrightarrow R(r,\theta)&=&\langle r\cos\theta,r\sin\theta, 4-r^2\rangle\\ \overrightarrow{R_r}(r,\theta)&=&\langle \cos\theta, \sin\theta, -2r\rangle\\ \overrightarrow{R_{\theta}}(r,\theta)&=&\langle -r\sin\theta,r\cos\theta,0\rangle\\ \overrightarrow{R_r}\times\overrightarrow{R_{\theta}}&=&\mbox{det }\left|\begin{array}{ccc} \overrightarrow i& \overrightarrow j& \overrightarrow k\newline \cos\theta & \sin\theta & -2r\newline -r\sin\theta& r\cos\theta & 0\end{array}\right|=\langle 2r^2\cos\theta, 2r^2\sin\theta,r\rangle. \end{eqnarray*} This vector is not consistent with the orientation because it is ``upward’’ (its last coordinate, also known as \(z\)-coordinate, is positive). In order to get the downward vector, we take \(\overrightarrow N=-\overrightarrow{R_r}\times\overrightarrow{R_{\theta}}= \langle -2r^2\cos\theta, -2r^2\sin\theta, -r\rangle\), hence \begin{eqnarray*} \iint_{\sigma} \overrightarrow F\cdot d\overrightarrow S&=&\iint_D \langle r\sin\theta, r\cos\theta, 4-r^2\rangle\cdot \langle -2r^2\cos\theta, -2r^2\sin\theta, -r\rangle\,drd\theta\\ &=&\int_0^{2}\int_0^{2\pi} \left(-2r^3\sin\theta\cos\theta-2r^3\sin\theta\cos\theta-4r+r^3\right)\,d\theta dr=2\pi\cdot \int_0^2\left(r^3-4r\right)\,dr=2\pi\cdot \left.\left(\frac{r^4}4-2r^2\right)\right|_0^2=-8\pi. \end{eqnarray*}

We have that \(\overrightarrow n\,dS=\pm\overrightarrow R_{s}\times\overrightarrow R_t\,ds dt\) where the sign will be chosen so that the obtained vector points away from \(z\) axis. \begin{eqnarray*} \overrightarrow R_s\times\overrightarrow R_t&=&\left\langle -9\sin s, 9\cos s, 0\right\rangle\times \left\langle 0,0,1\right\rangle\\ &=&\mbox{det }\left|\begin{array}{ccc}\overrightarrow i& \overrightarrow j& \overrightarrow k\\ -9\sin s & 9\cos s & 0\\ 0&0&1 \end{array}\right|\\ &=&\left\langle 9\cos s, 9\sin s,0\right\rangle. \end{eqnarray*} The obtained vector \(\overrightarrow R_s\times \overrightarrow R_t\) points away from \(z\) axis at any point because the \(x\) and \(y\) components of the vector at \((x_0,y_0,z_0)\) are precisely equal to \(x_0\) and \(y_0\). Thus the sign \(+\) should be chosen in place of \(\pm\). The integral now becomes: \begin{eqnarray*}\iint_G \overrightarrow F\cdot \overrightarrow n\,d S &=& \int_0^{2\pi}\int_0^{10}\left\langle 3,9\sin s,0 \right\rangle\cdot \left\langle 9\cos s,9\sin s,0\right\rangle\,dtds=I_1+I_2\;\mbox{ where }\\ I_1&=&\int_0^{2\pi}\int_0^{10}27\cos s\,dtds\\ \\I_2&=& \int_0^{2\pi}\int_0^{10}81\sin^2s\,dtds. \end{eqnarray*} We evaluate the integrals \(I_1\) and \(I_2\) in the following way: \begin{eqnarray*} I_1&=&\int_0^{2\pi}270\cos s\,ds=0. \\ I_2&=&\int_0^{2\pi} 810\sin^2 s\,ds=405\int_0^{2\pi}\left(1-\cos(2s)\right)\,ds=810\pi. \end{eqnarray*} We are now ready to evaluate the original integral: \begin{eqnarray*} \iint_G\overrightarrow F\cdot \overrightarrow n\,dS&=&I_1+I_2=810\pi. \end{eqnarray*}

The correct answer is E.

The expressions in options A and C doenot make sense as \(\nabla \overrightarrow F\) does not have a meaning. In option B \(\nabla f\) is a vector field. In option D the expression \(\nabla f\cdot\overrightarrow F\) is a scalar and \(\overrightarrow n\) is a vector.

We need to evaluate the integral \(\iint_G 1\,dS\), where \(G\) is the described surface. Let us first parametrize the surface. Our parameters will be \(r\) and \(\theta\) and the parametrization of the sphere above the \(xy\)-plane is: \begin{eqnarray*} x&=&r\cos\theta\\ y&=&r\sin\theta\\ z&=&\sqrt{100-r^2}, \end{eqnarray*} where \(r\) and \(\theta\) are real numbers such that \(0\leq r\leq 10\) and \(0\leq \theta\leq 2\pi\). The equation of the cylinder can be rewritten as \(x^2+y^2-10y=0\), or, equivalently \(r^2=10r\sin\theta\). The point \((x,y,z)\) is within the cylinder if and only if the corresponding \(r\) and \(\theta\) satisfy \(0\leq \theta\leq \pi\) and \(0\leq r\leq 10\sin\theta\). Thus the parametrization of \(G\) is: \[\overrightarrow R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,\sqrt{100-r^2}\right\rangle,\] for \(0\leq\theta\leq \pi\) and \(0\leq r\leq 10\sin\theta\). We have that \(dS=\left|\overrightarrow R_r\times\overrightarrow R_{\theta}\right|\,drd\theta\) hence we need to calculate \(\overrightarrow R_r\times\overrightarrow R_{\theta}\). \begin{eqnarray*} \overrightarrow R_r\times\overrightarrow R_{\theta}&=&\left\langle \cos\theta, \sin\theta, -\frac{r}{\sqrt{100-r^2}}\right\rangle\times \left\langle -r\sin\theta, r\cos\theta, 0\right\rangle\\ &=&\mbox{det }\left|\begin{array}{ccc}\overrightarrow i& \overrightarrow j& \overrightarrow k\\ \cos\theta & \sin\theta & -\frac{r}{\sqrt{100-r^2}}\\ -r\sin\theta & r\cos\theta & 0 \end{array}\right|\\ &=&\frac{r^2\cos\theta}{\sqrt{100-r^2}}\overrightarrow i + \frac{r^2\sin\theta}{\sqrt{100-r^2}}\overrightarrow j + r\overrightarrow k\\ &=&\frac{r}{\sqrt{100-r^2}}\left\langle r\cos\theta, r\sin\theta, \sqrt{100-r^2}\right\rangle. \end{eqnarray*} Therefore \[dS=\frac{r}{\sqrt{100-r^2}}\sqrt{r^2\cos^2\theta+r^2\sin^2\theta+100-r^2}\,drd\theta=\frac{10r}{\sqrt{100-r^2}}\,drd\theta.\] The integral now becomes \begin{eqnarray*} \iint_G1\,dS&=& \int_0^{\pi}\int_0^{10\sin\theta}\frac{10r}{\sqrt {100-r^2}}\,drd\theta. \end{eqnarray*} We will evaluate the inner integral using the substitution \(u=100-r^2\). Then we have \(du=-2r\,dr\) and as \(r\) goes from \(0\) to \(10\sin\theta\), the variable \(\theta\) would travel from \(100\) to \(100-100\sin^2\theta=100\cos^2\theta\). Therefore the integral becomes: \begin{eqnarray*} \int_0^{10\sin\theta}\frac{10r}{\sqrt{100-r^2}}\,dr&=&\int_{100}^{100\cos^2\theta}\frac{-5du}{\sqrt u}=5\int_{100\cos^2\theta}^{100} u^{-\frac12}\,du= \left.10u^{\frac12}\right|_{u=100\cos^2\theta}^{u=100}=100-100|\cos\theta|. \end{eqnarray*} We are now ready to evaluate the original integral: \begin{eqnarray*} \iint_G1\,dS&=&\int_0^{\pi} \left(100-100|\cos\theta|\right)|\,d\theta=200\pi-100\int_0^{\frac{\pi}2}\cos\theta\,d\theta-100\int_{\frac{\pi}2}^{\pi}(-\cos\theta)\,d\theta=200\pi-200. \end{eqnarray*}

The parametrization of the part of the paraboloid that is above the \(xy\)-plane is: \begin{eqnarray*} x&=&r\cos\theta\\ y&=&r\sin\theta\\ z&=&7-r^2, \end{eqnarray*} where \(r\) and \(\theta\) are real numbers such that \(0\leq \theta\leq 2\pi\) and \(0\leq r\leq \sqrt 7\). The parametrization can be written in the vector form as \[\overrightarrow R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,7-r^2\right\rangle,\quad \quad 0\leq\theta\leq 2\pi, \;\;0\leq r\leq \sqrt 7.\] We have that \(dS=\left|\overrightarrow R_r\times\overrightarrow R_{\theta}\right|\,drd\theta\) hence we need to calculate \(\overrightarrow R_r\times\overrightarrow R_{\theta}\). \begin{eqnarray*} \overrightarrow R_r\times\overrightarrow R_{\theta}&=&\left\langle \cos\theta, \sin\theta, -2r\right\rangle\times \left\langle -r\sin\theta, r\cos\theta, 0\right\rangle\\ &=&\mbox{det }\left|\begin{array}{ccc}\overrightarrow i& \overrightarrow j& \overrightarrow k\\ \cos\theta & \sin\theta & -2r\\ -r\sin\theta & r\cos\theta & 0 \end{array}\right|\\ &=&2r^2\cos\theta\overrightarrow i + 2r^2\sin\theta\overrightarrow j + r\overrightarrow k\\ &=&r\left\langle 2r\cos\theta, 2r\sin\theta, 1\right\rangle. \end{eqnarray*} Therefore \[dS=r\sqrt{4r^2\cos^2\theta+4r^2\sin^2\theta+1}\,drd\theta=r\sqrt{1+4r^2}\,drd\theta.\] The integral now becomes \begin{eqnarray*} \iint_G (x^2+y^2+2z)\,dS&=& \int_0^{2\pi}\int_0^{\sqrt 7}(14-r^2)\cdot r\sqrt{1+4r^2}\,drd\theta. \end{eqnarray*} We will evaluate the inner integral using the substitution \(u=1+4r^2\). Then we have \(du=8r\,dr\) and as \(r\) goes from \(0\) to \(\sqrt 7\), the variable \(u\) goes from \(1\) to \(29\). Therefore the integral becomes: \begin{eqnarray*} \int_0^{\sqrt 7}(14-r^2)\cdot r\sqrt{1+4r^2}\,dr&=&\frac18\int_{1}^{29} \left(14-\frac{u-1}4\right)\cdot\sqrt u\,du=\frac18\int_1^{29}\frac{57}4\sqrt u\,du-\frac1{32}\int_1^{29}u^{\frac32}\,du\\ &=&\left.\frac{57}{32}\cdot \frac23\cdot u^{\frac32}\right|_{u=1}^{u=29}-\left.\frac1{32}\cdot \frac25u^{\frac52}\right|_{u=1}^{u=29}\\ &=&\frac{957\sqrt{29}}{40}-\frac{47}{40}. \end{eqnarray*} We are now ready to evaluate the original integral: \begin{eqnarray*} \iint_G1\,dS&=&\int_0^{2\pi} \frac{957\sqrt{29}-47}{40}\,d\theta=\frac{957\sqrt{29}-47}{20}\pi. \end{eqnarray*}

Assume that \(M(x,y,z)\) is a point on the surface of revolution, and let \(\hat M (x,0,z)\) be its projection to the \(xz\) plane. Let us denote by \(\delta(M)\) its distance from the \(y\)-axis. Since the surface is obtained by rotating the curve around the \(y\)-axis, we must have \(y=f(\delta(M))\). We will parametrize the surface using parameters \(\delta\) and \(\theta\), where \(\delta\) is a distance of a point from the \(y\) axis, and \(\theta\) is the angle that the line \(\hat MO\) forms with the \(y\)-axis. The parametrization of the surface is \begin{eqnarray*} x&=&\delta\cos\theta\\ y&=&\delta^3+2\delta^2\\ z&=&\delta\sin\theta, \end{eqnarray*} where \(\delta\) and \(\theta\) are real numbers such that \(0\leq \theta\leq 2\pi\) and \(0\leq \delta\lneq 3\). The parametrization can be written in the vector form as \[\overrightarrow R(\delta,\theta)=\left\langle \delta\cos\theta,\delta^3+2\delta^2,d\sin\theta\right\rangle,\quad \quad 0\leq\theta\leq 2\pi, \;\;0\leq \delta\leq 3.\] We have that \(\overrightarrow n\,dS=\pm\overrightarrow R_{\delta}\times\overrightarrow R_{\theta}\,d\delta d\theta\) where the sign will be chosen so that the obtained vector forms an angle smaller than \(90^{\circ}\) with the \(y\) axis. Now we need to calculate \(\overrightarrow R_{\delta}\times\overrightarrow R_{\theta}\). \begin{eqnarray*} \overrightarrow R_{\delta}\times\overrightarrow R_{\theta}&=&\left\langle \cos\theta, 3\delta^2+4\delta, \sin\theta\right\rangle\times \left\langle -\delta\sin\theta, 0, \delta\cos\theta\right\rangle\\ &=&\mbox{det }\left|\begin{array}{ccc}\overrightarrow i& \overrightarrow j& \overrightarrow k\\ \cos\theta & 3\delta^2+4\delta & \sin\theta\\ -\delta\sin\theta & 0 &\delta\cos\theta \end{array}\right|\\ &=&\delta^2(3\delta+4)\cos\theta\overrightarrow i -\delta^2\sin\theta\overrightarrow j + \delta^2(3\delta+4)\sin\theta\overrightarrow k\\ &=&\delta^2\left\langle (3\delta+4)\cos\theta, -1, (3\delta+4)\sin\theta\right\rangle. \end{eqnarray*} The obtained vector \(\overrightarrow R_{\delta}\times \overrightarrow R_{\theta}\) has a negative dot product with the vector \(\langle 0,1,0\rangle\) corresponding to the \(y\) axis. That means that the angle it forms with the \(y\)-axis is greater than \(90^{\circ}\), therefore the sign \(-\) should be chosen and we get: \[\overrightarrow n\,dS=\delta^2\left\langle -(3\delta+4)\cos\theta, 1, -(3\delta+4)\sin\theta\right\rangle\,d\delta d\theta .\] The integral now becomes: \begin{eqnarray*}\iint_G \overrightarrow F\cdot \overrightarrow n\,d S &=& \int_0^{2\pi}\int_0^3\left\langle (\delta^3+2\delta^2)^2,\delta\cos\theta+\delta^3+2\delta^2,\delta\sin\theta\right\rangle\cdot \left\langle -(3\delta+4)\cos\theta, 1, -(3\delta+4)\sin\theta\right\rangle\delta^2\,d\delta d\theta=-I_1+I_2-I_3\;\mbox{ where }\\ I_1&=&\int_0^{2\pi}\int_0^3\left(\delta^3+2\delta^2\right)^2\cdot\delta^2\cdot(3\delta+4)\cos\theta\,d\delta d\theta\\ \\I_2&=& \int_0^{2\pi}\int_0^3\left(\delta\cos\theta+\delta^3+2\delta^2\right)\cdot\delta^2\,d\delta d\theta\\ \\ I_3&=& \int_0^{2\pi}\int_0^3(3\delta+4)\delta^3\sin^2\theta\,d\delta d\theta \end{eqnarray*} The first integral can be evaluated in the following way: \begin{eqnarray*}I_1&=&\int_0^{2\pi}\int_0^3\left(\delta^3+2\delta^2\right)^2\cdot\delta^2\cdot(3\delta+4)\cos\theta\,d\delta d\theta= \int_0^{3}\left(\delta^3+2\delta^2\right)^2\cdot\delta^2\cdot(3\delta+4)\,d\delta\cdot \int_0^{2\pi}\cos\theta\,d\theta=0. \end{eqnarray*} Let us now evaluate the second integral: \begin{eqnarray*} I_2&=&\int_0^{2\pi}\int_0^3\left(\delta\cos\theta+\delta^3+2\delta^2\right)\cdot\delta^2\,d\delta d\theta= \int_0^{2\pi}\int_0^3\delta^3\cos\theta\,d\delta d\theta+ \int_0^{2\pi}\int_0^3\left(\delta+2\right)\cdot\delta^4\,d\delta d\theta\\ &=& \int_0^3\delta^3\,d\delta\cdot \int_0^{2\pi} \cos\theta\,d\theta+2\pi\cdot \int_0^3\delta^5\,d\delta+4\pi\int_0^3\delta^4\,d\delta\\ &=& 0+2\pi\cdot \frac16\cdot 3^6+\frac{4\pi}5\cdot 3^5=\frac{3^7\pi}5. \end{eqnarray*} It remains to calculate the third integral: \begin{eqnarray*} I_3&=&\int_0^{2\pi}\int_0^3(3\delta+4)\delta^3\sin^2\theta\,d\delta d\theta=\int_0^{2\pi}\sin^2\theta\,d\theta\cdot \int_0^3\left(3\delta+4\right)\cdot\delta^3\,d\delta\\ &=&\int_0^{2\pi}\frac{1-\cos(2\theta)}2\,d\theta\cdot \int_0^3\left(3\delta^4+4\delta^3\right)\,d\delta=\pi\cdot \left( \frac35\cdot 3^5+3^4 \right)=\pi\cdot 3^4\cdot\frac{14}5\\&=&\frac{14\cdot\pi\cdot 3^4}5 \end{eqnarray*} We are now ready to evaluate the original integral: \begin{eqnarray*} \iint_G\overrightarrow F\cdot \overrightarrow n\,dS&=&-I_1+I_2-I_3=\frac{3^7\pi}5-\frac{14\cdot\pi\cdot 3^4}5=\frac{3^4\pi}5\cdot \left(3^3-14\right)=\frac{3^4\cdot 13\pi}5\\&=&\frac{1053\pi}5. \end{eqnarray*}