Evaluate the integral I = \int_S 4xyz \, dS where S is the part of the plane 2z = 2 - x +...


Evaluate the integral

{eq}I = \int_S 4xyz \, dS {/eq}

where {eq}S {/eq} is the part of the plane {eq}2z = 2 - x + 2y {/eq}

that lies above the rectangle {eq}[0,1 ] \times [0,1 ] {/eq} in the xy-plane.

Surface Integral:

Suppose that a surface is defined by {eq}z=g(x,y) {/eq} for

{eq}D\subset \mathbb{R}^2 {/eq}

The surface integral can then be computed as

{eq}\iint_S f(x,y,z)\, dS=\iint_D f(x,y,g(x,y))\sqrt{1+g_x^2+g_y^2}\, dA {/eq}

When the surface is defined using the variables in a different order, we use the same formula with the roles of the variables switched.

Answer and Explanation:

The surface {eq}2z=2-x+2y {/eq} can be rewritten as {eq}z=1-\frac{1}{2}x+y {/eq}. This gives us

{eq}z_x=\frac{-1}{2}\\ z_y=1 {/eq}

and thus

{eq}dS=\sqrt{1+(-1/2)^2+1^2}\, dA=\frac{3}{2}\, dA {/eq}

Therefore, for {eq}D=[0,1]\times [0,1] {/eq} in the {eq}xy {/eq}-plane we have

{eq}\begin{align} \iint_S 4xyz\, dS&=\iint_D 4xy(1-\frac{1}{2}x+y)\, dA\\ &=\int_0^1 \int_0^1 4xy-2x^2y+4xy^2\, dy\, dx\\ &=\int_0^1 2xy^2-x^2y^2+\frac{4}{3}xy^3|_0^1\, dx\\ &=\int_0^1 2x-x^2+\frac{4}{3}x\, dx\\ &=x^2-\frac{1}{3}x^3+\frac{2}{3}x^2|_0^1\\ &=1-\frac{1}{3}+\frac{2}{3}\\ &=\boxed{\frac{4}{3}} \end{align} {/eq}

Learn more about this topic:

Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14

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