# Evaluate the surface integral: Double integral over S of xz dS, where S is the boundary of the...

## Question:

Evaluate the surface integral: {eq}\iint_{S} xz \, \mathrm{d}S {/eq}, where {eq}S {/eq} is the boundary of the region enclosed by the cylinder {eq}y^2 + z^2 = 4 {/eq} and the planes {eq}x = 0 {/eq} and {eq}x + y = 6 {/eq}.

## Calculating a Surface Integral over a Parametrized Surface

Consider a surface {eq}S {/eq} parametrized by the vector-valued function

{eq}\mathbf{r}(u, v) = x(u, v) \mathbf{i} + y(u, v) \mathbf{j} + z(u, v) \mathbf{k} {/eq}

defined over a region {eq}D {/eq} in the {eq}uv{/eq}-plane. The surface integral of a function {eq}f(x, y, z) {/eq} over the surface {eq}S {/eq} is given by

{eq}\displaystyle\iint_S f(x, y, z) \: dS = \iint_D f(x(u, v), y(u, v), z(u, v)) ||\mathbf{r}_u(u, v) \times \mathbf{r}_v(u, v)|| \: dA {/eq}

In order to use this formula it is necessary to first parametrize the surface {eq}S {/eq} then calculate the cross product that appears in the integral. Then substitute the norm of the cross product along with the parametrized values of {eq}x, \: y, {/eq} and {eq}z {/eq} into the double integral and evaluate.

## Answer and Explanation:

The boundary of the region {eq}S {/eq} consists of three surfaces: the intersection of the cylinder with the {eq}yz{/eq}-plane, the cylinder itself,...

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