Evaluate the integral I = \int_S f \, dS where f(x,y,z) = z^2 + 3xy and S is the...

Question:

Evaluate the integral

{eq}I = \int_S f \, dS {/eq}

where {eq}f(x,y,z) = z^2 + 3xy {/eq} and {eq}S {/eq} is the portion of the plane

{eq}x + 2y + 2z = 0 {/eq}

above the unit disk

{eq}x^2 + y^2 \leq 1 {/eq}

in the xy-plane.

Surface Integrals:

A surface integral finds application in finding the area of the surface S, and is expressed as,

{eq}a(S) = \displaystyle \iint_S \; dS {/eq}

If the surface S can be expressed by the function, {eq}z = h(x,y) {/eq} then the surface integral can be transformed to an ordinary double integral as follows:

{eq}a(S) = \displaystyle \iint_S \; dS = \iint_R \sqrt{ 1 + z_x^2 + z_y^2}\; dA {/eq}

where {eq}R {/eq} is the projection of the surface S on to the xy - plane beneath {eq}S {/eq}.

We must evaluate the integral

{eq}I = \iint_S (z^2 + 3xy) \, dS {/eq}

where {eq}S {/eq} is the portion of the plane {eq}x + 2y + 2z = 0 {/eq} above the unit disk {eq}x^2 + y^2 \leq 1 {/eq} in the xy-plane.

{eq}\begin{align*} I &= \iint_S (z^2 + 3xy) \, dS \\ \end{align*} {/eq}

We will transform the above surface integral into an ordinary double integral, by projecting the surface {eq}S {/eq} onto the unit disk

{eq}x^2 + y^2 \leq 1 {/eq}

in the xy-plane.

The surface S is the plane, {eq}x + 2y + 2z = 0 {/eq}. From the equation of the plane we find the Jacobian of the transformation as follows:

{eq}\begin{align*} x + 2y + 2z &= 0 \\ 2z &= -x - 2y \\ z &= -\frac{x}{2} - y \\ &= -( \frac{x}{2} + y) \\ z_x &= -\frac{1}{2} \\ z_y &= -1\\ J_T &= \sqrt{1 + (-1)^2 + \bigg(\frac{-1}{2}\bigg)^2 } \\ &= \sqrt{1 + 1 + \frac{1}{4} } \\ &= \sqrt{2 + \frac{1}{4} } \\ &= \sqrt{\frac{9}{4} } \\ J_T&= \frac{3}{2} \end{align*} {/eq}

The surface integral becomes,

{eq}\begin{align*} I &= \iint_S (z^2 + 3xy) \, dS \\ &= \iint_R ( -( \frac{x}{2} + y) )^2 + 3xy \; J_T \; dA \\ &= \frac{3}{2} \iint_R ( -( \frac{x}{2} + y) )^2 + 3xy \; dA \\ &= \frac{3}{2} \iint_R ( \frac{x^2}{4} + y^2 + xy + 3xy ) \; dA\\ &= \frac{3}{2} \iint_R ( \frac{x^2}{4} + y^2 + 4xy ) \; dA\\ \end{align*} {/eq}

The region {eq}R {/eq} is the unit disk, which can be parametrized as,

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

The limits of integration are,

{eq}0 \leq \theta \leq 2 \pi \\ 0 \leq r \leq 1 {/eq}

The differential area element is given by,

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

The required integral becomes,

{eq}\begin{align*} I &= \iint_S (z^2 + 3xy) \, dS \\ &= \frac{3}{2} \iint_R \bigg( \frac{x^2}{4} + y^2 + 4xy \bigg) \; dA \\ &= \frac{3}{2} \int_0^{2\pi} \int_0^1 \bigg( \frac{r^2\cos^2 \theta }{4} + r^2\sin^2 \theta + 4r^2 \sin \theta \cos \theta \bigg) \; r \; dr \; d \theta \\ &= \frac{3}{8} \int_0^{2\pi} \int_0^1 ( \cos^2 \theta + 4 \sin^2 \theta + 16 \sin \theta \cos \theta ) \; r^3 \; dr \; d \theta \\ &= \frac{3}{8} \int_0^{2\pi} ( \cos^2 \theta + 4 \sin^2 \theta + 16 \sin \theta \cos \theta ) \bigg[ \frac{ r^4}{4} \bigg]_0^1 \; d \theta \\ &= \frac{3}{8} \int_0^{2\pi} ( \cos^2 \theta + 4 \sin^2 \theta + 16 \sin \theta \cos \theta ) \bigg[ \frac{ 1}{4} \bigg] \; d \theta \\ &= \frac{3}{32} \int_0^{2\pi} ( \cos^2 \theta + 4 \sin^2 \theta + 16 \sin \theta \cos \theta ) \; d \theta \\ &= \frac{3}{32} \int_0^{2\pi} ( \cos^2 \theta + 4 ( 1 - \cos^2\theta ) + 16 \sin \theta \cos \theta ) \; d \theta \\ &= \frac{3}{32} \int_0^{2\pi} ( \cos^2 \theta + 4 - 4 \cos^2\theta + 16 \sin \theta \cos \theta ) \; d \theta \\ &= \frac{3}{32} \int_0^{2\pi} ( 4 - 3 \cos^2\theta + 16 \sin \theta \cos \theta ) \; d \theta \\ &= \frac{3}{32} \int_0^{2\pi} \bigg( 4 - 3\bigg( \frac{ \cos 2\theta + 1}{2} \bigg)+ 16 \sin \theta \cos \theta \bigg) \; d \theta \\ &= \frac{3}{64} \int_0^{2\pi} \bigg( 8 - 3\bigg( \cos 2\theta + 1\bigg)+ 32\sin \theta \cos \theta \bigg) \; d \theta \\ &= \frac{3}{64} \int_0^{2\pi} \bigg( 5 - 3 \cos 2\theta + 32\sin \theta \cos \theta \bigg) \; d \theta \\ &= \frac{3}{64} \int \bigg[ 5\theta - \frac{3 \sin 2\theta}{2} + 16 \sin^2 \theta \bigg]_0^{2\pi} \\ &= \frac{3}{64} \int \bigg[ 5( 2 \pi) - \frac{3 \sin 4\pi }{2} + 16 \sin^2 ( 2 \pi) - \bigg( 5 \times 0 - \frac{3 \sin 0}{2} + 16 \sin^2 0 \bigg) \bigg] \\ &= \frac{3}{64} \times 10 \pi\\ &= \frac{15 \pi}{32} \end{align*} {/eq}

The required integral evaluates to,

{eq}\displaystyle I = \iint_S (z^2 + 3xy) \, dS = \frac{15 \pi}{32} {/eq}

Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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