Find the value of the surface integral \int \int_S 3yz dS, where S is part of the plane x + 3y +...

Question:

Find the value of the surface integral {eq}\int \int_S 3yz dS {/eq}, where {eq}S {/eq} is part of the plane {eq}x + 3y + 4z = 12 {/eq} that lies in the first octant.

Double Integral Across a Surface:

Double integrals are used to find the flux of a vector field through a given surface S. Here the function is scalar. So we have to find the limits of integration for {eq}x,y {/eq} with the help of expression {eq}x+3y+4z=12 {/eq} by putting z=0. Then evaluate the integral with integration rules

S is the surface given by {eq}x+3y+4z=12 {/eq}

Now {eq}\displaystyle z=3-\frac{x}{4}-\frac{3y}{4} {/eq}

The region is bounded by the points {eq}(12,0,0), (0,4,0) {/eq} and {eq}(0,0,3) {/eq}

Defining the limits of integration

{eq}\displaystyle 0 \leq y \leq 4-\frac{x}{3}, 0 \leq x \leq 12 {/eq}

Evaluating the integral

{eq}\displaystyle \int \int 3yzdS=\int_{0}^{12} \int_{0}^{4-\frac{x}{3}} 3y(3-\frac{x}{4}-\frac{3y}{4}) dydx {/eq}

{eq}\displaystyle \int \int 3yzdS=\int_{0}^{12} \int_{0}^{4-\frac{x}{3}} (9y-\frac{3xy}{4}-\frac{9y^2}{4})dydx {/eq}

{eq}\displaystyle \int \int 3yz dS=\int_{0}^{12} (\frac{9y^2}{2}-\frac{3xy^2}{8}-\frac{3y^3}{4})_{0}^{4-\frac{x}{3}} dx {/eq}

{eq}\displaystyle \int \int 3yz dS=\int_{0}^{12} [ 72+\frac{x^2}{2}-12x-6x-\frac{x^3}{24}+x^2-48+\frac{x^3}{36}+12x-x^2) ] dx {/eq}

{eq}\displaystyle \int \int 3yz dS=\int_{0}^{12} (24+\frac{x^2}{2}-6x-\frac{x^3}{72})dx {/eq}

{eq}\displaystyle \int \int 3yz dS=(24x-3x^2-\frac{x^4}{288})_{0}^{12} {/eq}

{eq}\displaystyle \int \int 3yz dS=288-432-72 {/eq}

{eq}\displaystyle \int \int 3yz dS= -216 {/eq}