# Chapter 16: Integrals and Vector Fields - Section 16.6 - Surface Integrals - Exercises 16.6 - Page 1000: 5

$3\sqrt 3$

#### Work Step by Step

Consider $\vec{r} (x,y) =x i+y j+(4-x-y) k$ So, $\vec{r_x} \times \vec{r_y}=i+j+k$ and $|\vec{r_x} \times \vec{r_y}| =\sqrt {1^2+1^2+1^2}=\sqrt 3$ Now, $\iint_{S} F (x,y,z) \ d \theta=\int_{0}^{1} \int_{0}^{1} (4-x-y) \times \sqrt 3 \ dy \ dx$ or, $=\sqrt 3 \int_{0}^{1} [4y-xy-\dfrac{y^2}{2} ]_0^1 \ dx$ or, $=\sqrt 3 \int_{0}^{1} 4(1-0)-x(1-0)-(\dfrac{1}{2}-0) \ dx$ or, $=\sqrt 3 \int_{0}^{1} (\dfrac{7}{2} -x)] \ dx$ By using a calculator, we evaluate the integral as: $\sqrt 3 \int_{0}^{1} (\dfrac{7}{2} -x)] \ dx= 3\sqrt 3$

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