1. Suppose the surface S is parameterized by \vec{r}(u,v) = \langle u +v, 3u^2, u-v \rangle, \...

Question:

Suppose the surface {eq}S {/eq} is parameterized by {eq}r(u,v) =(u +v, 3u^2, u-v), \quad 0\leq u \leq 2, -1\leq v \leq 1 {/eq}

(a) Find the equation of the plane tangent to {eq}S {/eq} at the point {eq}r(1,1) {/eq}.

(b) Set up the integral needed to compute the surface area of {eq}S {/eq}.

(c) Set up the integral to integrate the function {eq}f(x,y,z) = x + z -\sqrt{3y} {/eq} along {eq}S {/eq}.

(d) Set up the integral to compute the flux of {eq}\vec{F} = \langle 3z, -y^2, 3x \rangle {/eq} across {eq}S {/eq}.

Vector and Scalar Surface Integrals:

Suppose that {eq}S {/eq} is an oriented surface in {eq}\mathbb{R}^3 {/eq}, which is parameterized by some function {eq}r(u,v)=(x(u,v),y(u,v),z(u,v)) {/eq}, whose domain is some region {eq}R {/eq} in the {eq}uv {/eq}-plane. There are two different notions of integration over {eq}S {/eq}:

1) If {eq}f(x,y,z) {/eq} is any function whose domain includes {eq}S {/eq}, then the scalar surface integral of {eq}f {/eq} over {eq}S {/eq} is defined to be

{eq}\displaystyle \iint_S f \, dS = \iint_R f(r(u,v)) \|r_u \times r_v\| \, dA \, . {/eq}

2) If {eq}\vec{F}(x,y,z) {/eq} is any vector field whose domain includes {eq}S {/eq}, then the vector surface integral of {eq}\vec{F} {/eq} over {eq}S {/eq} is defined to be

{eq}\displaystyle \iint_S \vec{F} \cdot \vec{dS}=\iint_R \vec{F}(r(u,v)) \cdot (r_u \times r_v) \, dA \, . {/eq}

The vector surface integral of {eq}\vec{F} {/eq} over {eq}S {/eq} is sometimes also called the flux of {eq}\vec{F} {/eq} over {eq}S {/eq}.

Answer and Explanation:

We'll start by making a few computations that will come in handy later. Taking the partial derivatives of the parameterization {eq}r(u,v) =(u +v, 3u^2, u-v) {/eq} gives:

{eq}\begin{align*} r_u&=\left<1,6u,1\right>\\ r_v&=\left<1,0,-1\right> \, . \end{align*} {/eq}

The cross product of these partial derivatives gives a positively oriented normal vector to {eq}S {/eq}. It is:

{eq}\begin{align*} r_u \times r_v&=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 6u & 1\\ 1 & 0 & -1 \end{vmatrix}&\text{(by the determinant mnemonic for the cross product)}\\ &=(6u(-1)-1(0))\mathbf{i}+(1(1)-1(-1))\mathbf{j}+(1(0)-6u(1))\mathbf{k}\\ &=-6u\mathbf{i}+2\mathbf{j}-6u\mathbf{k} \, . \end{align*} {/eq}

Finally, the magnitude of this cross product is:

{eq}\begin{align*} \|r_u \times r_v\| \, dv \, du&= \|\left<-6u,2,-6u\right>\|&\text{(by the above cross product computation)}\\ &=\sqrt{(-6u)^2+2^2+(-6u)^2}&\text{(by the definition of vector magnitude)}\\ &=\sqrt{36u^2+4+36u^2} \\ &=\sqrt{72u^2+4} \, . \end{align*} {/eq}

(a) First, the point {eq}r(1,1) {/eq} has the coordinates:

{eq}\begin{align*} r(1,1)&=(1+1,3(1^2),1-1)\\ &=(1,3,0) \, . \end{align*} {/eq}

Next, we find a normal vector to {eq}r(1,1) {/eq}:

{eq}\begin{align*} \left.(r_u \times r_v)\right|_{(1,1)}&=\left<-6(1),2,-6(1)\right>&\text{(by the above computation of the cross product)}\\ &=\left<-6,2,-6\right>\, . \end{align*} {/eq}

The tangent plane to {eq}S {/eq} at {eq}r(1,1) {/eq} will be the plane of points whose displacement vector to {eq}r(1,1)=(1,3,0) {/eq} is orthogonal to {eq}\left.(r_u \times r_v)\right|_{(1,1)}=\left<-6,2,-6\right> {/eq}. So, if {eq}(x,y,z) {/eq} is in that plane, then:

{eq}\begin{align*} \left<-6,2,-6\right> \cdot \left<x-1,y-3,z-0\right>&=0\\ \left<-6,2,-6\right> \cdot \left<x-1,y-3,z\right>&=0\\ -6(x-1)+2(y-3)-6z&=0&\text{(evaluating the dot product)}\\ -6x+6+2y-6-6z&=0&\text{(expanding)}\\ -6x+2y-6z&=0 \, . \end{align*} {/eq}

That is, the desired tangent plane has the equation {eq}\boxed{-6x+2y-6z=0} \, {/eq}.

(b) The surface area of {eq}S {/eq} is computed by the scalar surface integral of the constant function 1 over {eq}S {/eq}. That is, the surface area of {eq}S {/eq} can be found as follows:

{eq}\begin{align*} \iint_S 1 \, dS &= \int_{u=0}^2 \int_{v=-1}^1 \|r_u \times r_v\| \, dv \, du&\text{(by the definition of the scalar surface integral)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 \sqrt{72u^2+4} \,dv \, du \&\text{(by the above computation of }\|r_u \times r_v\|\text{).} \end{align*} {/eq}

So the surface area of {eq}S {/eq} is computed by the iterated integral {eq}\boxed{\int_{u=0}^2 \int_{v=-1}^1 \sqrt{72u^2+4} \,dv \, du \, .} {/eq}

(c) To integrate the function {eq}f(x,y,z)=x+z-\sqrt{3y} {/eq} over {eq}S {/eq}, we apply the definition of the scalar surface integral:

{eq}\begin{align*} \iint_S f(x,y,z) \, dS &= \int_{u=0}^2 \int_{v=-1}^1 f(r(u,v)) \|r_u \times r_v\| \, dv \, du&\text{(by the definition of the scalar surface integral)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 f(u+v,3u^2,u-v)\sqrt{72u^2+4} \, dv \, du&\text{(by the definition of }r(u,v)\text{, and the norm of its cross product which we computed above)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 (u+v+u-v-\sqrt{3(3u^2)})\sqrt{72u^2+4} \, dv \, du\\ &=\int_{u=0}^2 \int_{v=-1}^1 (2u-\sqrt{9u^2})\sqrt{72u^2+4} \, dv \, du&\text{(canceling, multiplying)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 (2u-3u)\sqrt{72u^2+4} \, dv \, du&\text{(evaluating the square root; note that }u \ge 0\text{ on the domain of integration)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 -u\sqrt{72u^2+4} \, dv \, du&\text{(combining like terms).} \end{align*} {/eq}

So we can compute the integral {eq}\displaystyle \iint_S (x + z - \sqrt{3y}) \, dS {/eq} by evaluating the iterated integral {eq}\boxed{\int_{u=0}^2 \int_{v=-1}^1 -u\sqrt{72u^2+4} \, dv \, du \, .} {/eq}

(d) The flux of {eq}\vec{F}(x,y,z)=\left<3z,-y^2,3x\right> {/eq} across {eq}S {/eq} is given by the following vector surface integral:

{eq}\begin{align*} \int_S \vec{F} \cdot \vec{dS}&= \int_{u=0}^2 \int_{v=-1}^1 \vec{F}(r(u,v)) \cdot (r_u \times r_v) \, dv \, du&\text{(by the definition of the vector surface integral)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 \vec{F}(u+v,3u^2,u-v) \cdot \left<-6u, 2, -6u\right> \, dv \, du&\text{(by the above expressions for }r\text{ and }r_u \times r_v\text{)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 \left<3(u-v),-(3u^2)^2,3(u+v)\right>\cdot \left<-6u,2,-6u\right> \, dv \, du&\text{(by the given expression for }vec{F}\text{)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 \left<3u-3v,-9u^4,3u+3v\right> \cdot \left<-6u,2,-6u\right> \, dv \, du&\text{(simplifying the first factor)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 \left[(3u-3v)(-6u)+(-9u^4)(2)+(3u+3v)(-6u)\right] \, dv \, du&\text{(evaluating the dot product)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 \left(-18u^2+18uv-18u^4-18u^2-18uv\right) \, dv \, du&\text{(expanding)}\\ &=\int_{u=0}^2 \int_{v=-1}^1 \left(-36u^2-18u^4\right) \, dv \, du&\text{(combining/canceling like terms).} \end{align*} {/eq}

So the flux of {eq}\vec{F} {/eq} through {eq}S {/eq} is computed by the iterated integral {eq}\boxed{\int_{u=0}^2 \int_{v=-1}^1 \left(-36u^2-18u^4\right) \, dv \, du \, .} {/eq}


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Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
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