Use Green's Theorem to evaluate the line integral along the positively oriented closed curve $C$. $\oint _ { C } 3 x y d x + 4 x ^ { 2 } d y$, where $\mathrm { C }$ is the triangle with vertices $( 0,0 ) , ( 3,4 )$, and $( 0,4 )$
A) 6
B) 60
C) 30
D) 66

Question

Quizplus · Accepted Answer

Green's Theorem states that if $\mathrm{M}(x,y)$ and $\mathrm{N}(x,y)$ are continuously differentiable functions on a simply connected region $\mathrm{D}$ bounded by a piecewise smooth, simple, closed curve $\mathrm{C}$, then 
$$\oint _ { C } \left[ \mathrm { M } ( x , y ) d x + \mathrm { N } ( x , y ) d y \right] = \iint _ { D } \left( \frac { \partial \mathrm { N } } { \partial x } - \frac { \partial \mathrm { M } } { \partial y } \right) d A.$$
Here, we have 
$$\mathrm M(x,y)=4x^2 \quad \mathrm{and} \quad \mathrm N(x,y)=3xy.$$
Taking partial derivatives, we get 
$$\frac{\partial \mathrm{N}}{\partial x}=3y \quad \mathrm{and} \quad \frac{\partial \mathrm{M}}{\partial y}=0.$$
Hence, by Green's Theorem, we have
\begin{align*}
\oint _ { C } \left[ \mathrm { M } ( x , y ) d x + \mathrm { N } ( x , y ) d y \right] &= \iint _ { D } \left( \frac { \partial \mathrm { N } } { \partial x } - \frac { \partial \mathrm { M } } { \partial y } \right) d A \
&= \iint _{D} 3y \ d A \
&= \int_{0}^{3} \int_{0}^{y/3} 3y \ dx dy \
&= \int_{0}^{3} \left[3xy\right]_{x=0}^{x=y/3} dy \
&= \int_0^3 (y^2) dy \
&= \left[ \frac{y^3}{3} \right]^{3}_{0} \
&= 9.
\end{align*}
Since the curve is closed, this value is also equal to the line integral around the boundary of the region $\mathrm{D}$, which is the triangle with vertices $(0,0), (3,4),$ and $(0,4)$.

Therefore, the answer is \boxed{\textbf{(C) } 30}.

Use Green's Theorem to Evaluate the Line Integral Along the Positively