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13.3 - Think before you compute
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<h2>
[13.3] - Think before you compute
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<ol>
<li>Compute $\int_C (ye^{xy}\uv i + xe^{xy}\uv j)\cdot d\myv r$ for the curve $C$ shown below.<br>
<img src="0g/13.3.crazy1.jpg"><br>
<p class="result justme">
A potential function such that $\myv \grad f=ye^{xy}\uv i + xe^{xy}\uv j)$ is $f(x,y)=e^{xy}$.  Therefore,
the integral is:
$$\int_{\myv r_i}^{\myv r_f} \myv\grad f\cdot d\myv r =f(\myv r_f)-f(\myv r_i),$$
where the curve runs from $\myv r_i=\langle -1,-1\rangle$ to $\myv r_f=\langle 3,1\rangle$.  So the answer is
$e^{3*1}-e^{-1*-1}$<span class="answer">$=e^3-e^1$<span>


<li style="margin-top: 26ex">
Compute $\int_C (yz^2\uv i + xz^2\uv j+2xyz\uv k)\cdot d\myv r$ for the curve $C$ shown below.<br>
<img src="0g/13.3.crazy2.jpg"><br>
<p class="result justme">
A potential function that gives this vector field is $f(x,y,z)=xyz^2$.  So the integral is $f(\myv r_f)-f(\myv r_i)=f(0,1,2)-f(2,0,0)=(0*1*2^2)-(2*0*0)$<span class="answer">$=0$</span>.

<li style="page-break-before: always">
Let $\myv F(x,y)=(2x+y)\cos(x^2+xy)\uv i+\left(x\cos(x^2+xy)+1\right)\uv j$.
<ol type="a">
<li>Show that $\myv F$ is a conservative vector field.
<p class="result">
$$\begineq
\frac{\del P}{\del y}&=&\frac{\del }{\del y}(2x+y)\cos(x^2+xy)\\
&=&1*\cos(x^2+xy) +(2x+y)\sin(x^2+xy)*x\\
\endeq$$

$$\begineq
\frac{\del Q}{\del x}&=&\frac{\del }{\del x}(x\cos(x^2+xy)+1)\\
&=&\cos(x^2+xy) +x\sin(x^2+xy)*(2x+y)
\endeq$$
<span class="answer">Since these are equal</span>, $\myv F$ is a conservative vector field.
<li style="margin-top: 26ex">
Let $C$ be the curve parameterized by $\myv r(t)=\sin t\uv i+(1-\cos t)\uv j$
with $0\leq t\leq \pi$.<br>
Find $\int_C \myv F \cdot d\myv r$.
<p class="result">
$\myv r_i=\myv r(0)=\sin(0)\uv i +(1-\cos(0))\uv j=(0,0)$, and
$\myv r_f=\myv r(\pi)=\sin(\pi)\uv i +(1-\cos(\pi))\uv j=(0,2)$.
<br><br>
A potential function is $f(x,y)=-\sin(x^2-2y)$.  So the integral is
$f(0,2)-f(0,0)=-\sin(0^2-4)-(-\sin(0^2-2*0))=-\sin(-4)$
<span class="answer">= -0.757</span>.
</ol>
</ol>
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