Evaluate the integral I = \frac{1}{4}\int_S dS where S is the surface given parametrically...

Question:

Evaluate the integral

{eq}I = \frac{1}{4}\int_S dS {/eq}

where {eq}S {/eq} is the surface given parametrically by

{eq}\mathbf{ \phi}(u,v) = (2uv, u+v, u - v) {/eq}

for {eq}u^2 + v^2 \leq 4 {/eq}

Surface Integrals:

There are a number of ways to evaluate a surface integral. Often the most straightforward way involves using a parameterization of the surface. Then we can use the following integral formula:

{eq}\begin{align*} \iint_S f(x,y,z)\ dS ****= \iint_D f(u,v) | \vec r_u \times \vec r_v |\ dA \end{align*} {/eq}

Note that we will need the partials of the parameterization. We will be using polar coordinates:

{eq}x = r \cos \theta {/eq}

{eq}y = r \sin \theta {/eq}

{eq}r^2 = x^2+y^2 {/eq}

{eq}\theta = \tan^{-1} \frac{y}{x} {/eq}

{eq}dA = r\ dr\ d\theta {/eq}

Answer and Explanation:

Note that we are effectively integrating {eq}\frac14 {/eq}, so what we are computing is a quarter of the surface area of the surface that lies above the circular region. We have

{eq}\begin{align*} \vec r_u &= \frac{d}{du} \left< 2uv, u+v, u-v \right> \\ &= \left< 2v, 1, 1 \right> \end{align*} {/eq}

and

{eq}\begin{align*} \vec r_v &= \frac{d}{dv} \left< 2uv, u+v, u-v \right> \\ &= \left< 2u, 1, -1 \right> \end{align*} {/eq}

Their cross product is

{eq}\begin{align*} \vec r_u \times \vec r_v &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2v & 1 & 1 \\ 2u & 1 & -1 \end{vmatrix} \\ &= \left< -1-1,\ -(-2v-2u),\ 2v-2u \right> \\ &= 2 \left< -1,\ v+u,\ v-u \right> \end{align*} {/eq}

which has magnitude

{eq}\begin{align*} | \vec r_u \times \vec r_v | &= | 2 \left< -1,\ v+u,\ v-u \right> | \\ &= 2 \sqrt{\left( -1 \right)^2+\left( v+u \right)^2+\left( v-u \right)^2} \\ &= 2 \sqrt{1 + v^2 + 2uv + u^2 + v^2 - 2uv +u^2} \\ &= 2 \sqrt{1 + 2(u^2 + v^2)} \\ &= 2 \sqrt{1+2r^2} \end{align*} {/eq}

where we switched to polar coordinates as soon as it was convenient. Now, the region is {eq}r \leq 2 {/eq} when {eq}\theta \in [0,2\pi] {/eq}, so we find

{eq}\begin{align*} \frac14 \iint_S\ dS &= \frac14 \int_0^{2\pi} \int_0^2 \left( 2 \sqrt{1+2r^2} \right)\ r\ dr\ d\theta \\ &= \frac12 \int_0^{2\pi} d \theta \int_0^2 r \sqrt{1+2r^2}\ dr \\ &= \frac12 (2\pi) \left [ \frac16(1+2r^2)^{3/2} \right ]_0^2 \\ &= \frac\pi6 \left[ (1+2(2)^2)^{3/2} - (1+0)^{3/2} \right] \\ &= \frac{13\pi}{3} \\ &\approx 13.6136 \end{align*} {/eq}


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Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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