Evaluate the integral along the path C defined as
$y = 9 - x ^ { 2 } \text { from } ( 0,3 ) \text { to } ( 3,0 ) \text {. }$
A) $- \frac { 1755 } { 7 }$
B) $- \frac { 303 } { 2 }$
C) $- \frac { 909 } { 7 }$
D) $- \frac { 1323 } { 8 }$
E) $- \frac { 441 } { 2 }$

Question

Evaluate the integral  along the path  C  defined as  
$y = 9 - x ^ { 2 } \text { from } ( 0,3 ) \text { to } ( 3,0 ) \text {. }$
A) $- \frac { 1755 } { 7 }$
B) $- \frac { 303 } { 2 }$
C) $- \frac { 909 } { 7 }$
D) $- \frac { 1323 } { 8 }$
E) $- \frac { 441 } { 2 }$

Quizplus · Accepted Answer

We first need to parameterize the path $C$. We can do this by setting $x=t$ and $y=9-t^2$, where $t$ is the parameter running from $0$ to $3$. Then, we have
\begin{align*}
\int_C x^2y\, dx + (3y+x^2)\, dy &= \int_0^3 t^2(9-t^2)\cdot 1\, dt + (3(9-t^2)+t^2)(-2t)\, dt \
&= \int_0^3 (9t^2-t^4)\, dt + (-6t^3+27t)\, dt \
&= \left[\frac{9}{3}-\frac{1}{5}\cdot 3^5ight]+(-0+27\cdot 3)-\left[0-\frac{9}{2}ight] \
&= -\frac{441}{2}.
\end{align*}
Therefore, our answer is $\boxed{	ext{(E) } -\frac{441}{2}}$.

Evaluate the Integral Along the Path C y=9−x2 from (0,3) to (3,0). y = 9 - x ^ { 2 } \text { from } ( 0,3 ) \text { to } ( 3,0 ) \text {. }y=9−x2 from (0,3) to (3,0).

Evaluate the Integral Along the Path C $y = 9 - x ^ { 2 } \text { from } ( 0,3 ) \text { to } ( 3,0 ) \text {. }$