The surface area of the surface z = xy in the first octant lying inside $x ^ { 2 } + y ^ { 2 } = 1$ is

A) $\frac { \pi ( 2 \sqrt { 2 } - 1 ) } { 3 }$
B) $\frac { \pi ( 2 \sqrt { 2 } - 1 ) } { 6 }$
C) $\frac { \pi ( 7 \sqrt { 21 } - 9 ) } { 6 }$
D) $\frac { \pi ( 7 \sqrt { 21 } - 1 ) } { 3 }$
E) $\frac { \pi ( 2 \sqrt { 2 } - 1 ) } { 2 }$

Question

The surface area of the surface z = xy in the first octant lying inside $x ^ { 2 } + y ^ { 2 } = 1$ is

A) $\frac { \pi ( 2 \sqrt { 2 } - 1 ) } { 3 }$
B) $\frac { \pi ( 2 \sqrt { 2 } - 1 ) } { 6 }$
C) $\frac { \pi ( 7 \sqrt { 21 } - 9 ) } { 6 }$
D) $\frac { \pi ( 7 \sqrt { 21 } - 1 ) } { 3 }$
E) $\frac { \pi ( 2 \sqrt { 2 } - 1 ) } { 2 }$

Quizplus · Accepted Answer

The surface area is calculated using the formula for surface area of a parametric surface. For $z = xy$, the integral over the region $x^2 + y^2 \leq 1$ in the first octant gives the result $\frac{\pi (2\sqrt{2} - 1)}{6}$.

The Surface Area of the Surface Z = Xy in the First