Find the area of the surface.
The part of the paraboloid $\mathbf { r } ( u , v ) = u \cos v \mathbf { i } + u \sin v \mathbf { j } + u ^ { 2 } \mathbf { k } ; 0 \leq u \leq 3,0 \leq v \leq 2 \pi$
A) $\frac { 5 \pi ( 37 \sqrt { 37 } - 1 ) } { 12 }$
B)
$\frac { \pi ( 37 \sqrt { 37 } - 1 ) } { 6 }$
C) $\frac { \pi ( 37 \sqrt { 37 } - 1 ) } { 12 }$
D) $\frac { \pi ( 37 \sqrt { 37 } - 1 ) } { 3 }$

Question

Quizplus · Accepted Answer

The surface area of the given paraboloid can be found by evaluating the integral of the magnitude of the cross product of the partial derivatives of $\mathbf{r}(u, v)$ with respect to $u$ and $v$. After computing, the correct result is $\frac{\pi (37 \sqrt{37} - 1)}{6}$.

Find the Area of the Surface r(u,v)=ucos⁡vi+usin⁡vj+u2k;0≤u≤3,0≤v≤2π\mathbf { r } ( u , v ) = u \cos v \mathbf { i } + u \sin v \mathbf { j } + u ^ { 2 } \mathbf { k } ; 0 \leq u \leq 3,0 \leq v \leq 2 \pir(u,v)=ucosvi+usinvj+u2k;0≤u≤3,0≤v≤2π

Find the Area of the Surface $\mathbf { r } ( u , v ) = u \cos v \mathbf { i } + u \sin v \mathbf { j } + u ^ { 2 } \mathbf { k } ; 0 \leq u \leq 3,0 \leq v \leq 2 \pi$