The surface integral $\int \int$S $y d S$ where S is $z = 4 - y ^ { 2 }$ in the first octant bounded by $x = 3$ and the coordinate planes, is

A) $\frac { 17 \sqrt { 17 } - 1 } { 2 }$
B) $\frac { 17 \sqrt { 17 } + 1 } { 2 }$
C) $\frac { 17 \sqrt { 17 } - 1 } { 4 }$
D) $\frac { 17 \sqrt { 17 } + 1 } { 4 }$
E) $\frac { 17 \sqrt { 17 } - 1 } { 3 }$

Question

The surface integral $\int \int$S $y d S$ where S is $z = 4 - y ^ { 2 }$ in the first octant bounded by $x = 3$ and the coordinate planes, is

A) $\frac { 17 \sqrt { 17 } - 1 } { 2 }$
B) $\frac { 17 \sqrt { 17 } + 1 } { 2 }$
C) $\frac { 17 \sqrt { 17 } - 1 } { 4 }$
D) $\frac { 17 \sqrt { 17 } + 1 } { 4 }$
E) $\frac { 17 \sqrt { 17 } - 1 } { 3 }$

Quizplus · Accepted Answer

The Answer of The surface integral $\int \int$S $y d S$ where S is $z = 4 - y ^ { 2 }$ in the first octant bounded by $x = 3$ and the coordinate planes, is

A) $\frac { 17 \sqrt { 17 } - 1 } { 2 }$
B) $\frac { 17 \sqrt { 17 } + 1 } { 2 }$
C) $\frac { 17 \sqrt { 17 } - 1 } { 4 }$
D) $\frac { 17 \sqrt { 17 } + 1 } { 4 }$
E) $\frac { 17 \sqrt { 17 } - 1 } { 3 }$

The Surface Integral ∫∫\int \int∫∫ S ydSy d SydS Where S Is z=4−y2z = 4 - y ^ { 2 }z=4−y2

The Surface Integral $\int \int$ S $y d S$ Where S Is $z = 4 - y ^ { 2 }$