## Related questions with answers

Find the volume of a region

$\mathcal { W }$

if

$\iint _ { \partial \mathcal { W } } \left\langle x + x y + z , x + 3 y - \frac { 1 } { 2 } y ^ { 2 } , 4 z \right\rangle \cdot d \mathbf { S } = 16$

Solution

VerifiedGiven a vector field $\textbf{F}$ $\bigg \langle x+xy+z, x+3y-\dfrac{1}{2}y^2, 4z \bigg \rangle$ from the integral. We solve for the volume of the region $\mathcal{W}$ using the information given in the Divergence Theorem.

$\begin{align} \int \hspace{-3mm} \int_{\mathcal{S}} \textbf{F} \cdot \,d\textbf{S}&= \int \hspace{-3mm} \int_{\mathcal{W}} \text{div}(\textbf{F}) \,dV=16 \end{align}$

Solving for the curl, we have the following:

$\begin{align*} \text{div}(\textbf{F})&= \dfrac{\partial}{\partial x} (x+xy+z) + \dfrac{\partial}{\partial y} \left(x+3y-\dfrac{y^2}{2}\right)+\dfrac{\partial}{\partial z} (4z)\\ &=1+y+3-y=8 \end{align*}$

From (1), we get the following equality:

$\begin{align*} \int \hspace{-3mm} \int_{\mathcal{S}} \textbf{F} \cdot \,d\textbf{S}&= \int \hspace{-3mm} \int_{\mathcal{W}} (8) \,dV= 8 \int \hspace{-3mm} \int_{\mathcal{W}} (1) \,dV=8 \text{Volume}(\mathcal{W})=16 \end{align*}$

Thus, from the equation above, we get the the volume of the region $\mathcal{W}$ to be:

$\begin{align*} \text{Volume}(\mathcal{W})=2 \end{align*}$

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