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Question

Find the volume of a region

W\mathcal { W }

if

Wx+xy+z,x+3y12y2,4zdS=16\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

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Answered 2 years ago
Answered 2 years ago
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Given a vector field F\textbf{F} x+xy+z,x+3y12y2,4z\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 W\mathcal{W} using the information given in the Divergence Theorem.

SFdS=Wdiv(F)dV=16\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:

div(F)=x(x+xy+z)+y(x+3yy22)+z(4z)=1+y+3y=8\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:

SFdS=W(8)dV=8W(1)dV=8Volume(W)=16\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 W\mathcal{W} to be:

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

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