Evaluate the surface integral of G over the surface S.
-$S$ is the parabolic cylinder $y = 3 x ^ { 2 } , 0 \leq x \leq 2$ and $0 \leq z \leq 4 ; G ( x , y , z ) = 3 x$
A) $\frac { 1 } { 9 } ( 145 \sqrt { 145 } - 1 )$
B) $\frac { 1 } { 3 } ( 145 \sqrt { 145 } - 1 )$
C) $\frac { 1 } { 9 } ( 145 \sqrt { 145 } + 1 )$
D) $\frac { 1 } { 3 } ( 145 \sqrt { 145 } + 1 )$

Question

Quizplus · Accepted Answer

Explanation:The surface integral of G over the surface S is given by$$\iint_S G(x, y, z) dS = \int_0^2 \int_0^4 3x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dz dx$$$$= \int_0^2 \int_0^4 3x \sqrt{1 + (6x)^2} dz dx$$$$= \int_0^2 \left[ 3xz\sqrt{1 + 36x^2} \right]_0^4 dx$$$$= \int_0^2 12x\sqrt{1 + 36x^2} dx$$$$= \left[ \frac{1}{18} (1 + 36x^2)^{3/2} \right]_0^2$$$$= \frac{1}{18} (145\sqrt{145} - 1)$$

Evaluate the Surface Integral of G Over the Surface S SSS

Evaluate the Surface Integral of G Over the Surface S $S$