$\text { Evaluate } \iint _ { S } ( x , y , z ) d S \text {, where } f ( x , y , z ) = \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \text { and } S \text { is given by }$ $z = \sqrt { x ^ { 2 } + y ^ { 2 } } , x ^ { 2 } + y ^ { 2 } \leq 121$
A) $\frac { 5324 } { 5 \pi }$
B) $\frac { 2662 \pi } { 3 }$
C) $\frac { 5324 \pi } { 3 }$
D) $\frac { 1331 \pi } { 5 }$
E) $\frac { 2662 } { 3 \pi }$

Question

Quizplus · Accepted Answer

The surface $S$ described by $z = \sqrt{x^2 + y^2}$ is a cone. The function $f(x, y, z) = \sqrt{x^2 + y^2 + z^2}$ represents the distance from the origin to a point $(x, y, z)$ on the surface. Given the boundary $x^2 + y^2 \leq 121$, the surface integral calculates the total "weighted" surface area of the cone, where the "weight" is the distance from the origin. To solve this, we can switch to cylindrical coordinates, where $x^2 + y^2 = r^2$, $z = r$, and $dS = \sqrt{1 + (\frac{dz}{dr})^2}rdrd\theta$. The integral becomes $\int_0^{2\pi} \int_0^{11} \sqrt{r^2 + r^2} \sqrt{1 + 1} rdrd\theta$, which simplifies to $\int_0^{2\pi} \int_0^{11} 2r^2 drd\theta$. Solving this gives $\frac{5324\pi}{3}$.

z=x2+y2,x2+y2≤121z = \sqrt { x ^ { 2 } + y ^ { 2 } } , x ^ { 2 } + y ^ { 2 } \leq 121z=x2+y2​,x2+y2≤121

$z = \sqrt { x ^ { 2 } + y ^ { 2 } } , x ^ { 2 } + y ^ { 2 } \leq 121$