Evaluate $\iint _ { S } f ( x , y , z ) d S$.
$f ( x , y , z ) = z ; S$ is the part of the torus with vector representation
$\mathbf { r } ( u , v ) = ( 5 + 3 \cos v ) \cos u \mathbf { i } + ( 5 + 3 \cos v ) \sin u \mathbf { j } + 3 \sin v \mathbf { k } , 0 \leq u \leq 2 \pi , 0 \leq v \leq \frac { \pi } { 2 }$
A) 0
B) $117 \pi$
C) 117
D) $27 \pi$

Question

Quizplus · Accepted Answer

The given function $f(x, y, z) = z$ and the surface $S$ is a part of a torus. The vector representation of the torus is given by $\mathbf{r}(u, v)$. To find the surface integral $\iint_{S} f(x, y, z) dS$, we substitute $z$ with its representation in terms of $u$ and $v$, which is $3\sin v$. The surface element $dS$ can be found by computing the magnitude of the cross product of the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$, denoted as $\left|\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\right|$. After integrating $3\sin v$ over the given limits for $u$ and $v$, taking into account the Jacobian (the magnitude of the cross product), the result is $117\pi$, which corresponds to the total area of the specified part of the torus multiplied by the function $f(x, y, z) = z$.

Evaluate ∬Sf(x,y,z)dS\iint _ { S } f ( x , y , z ) d S∬S​f(x,y,z)dS

Evaluate $\iint _ { S } f ( x , y , z ) d S$