Using Green's Theorem, the line integral $\int $C $x ^ { 2 } y d x - y ^ { 2 } x d y$ where C is the circle $x ^ { 2 } + y ^ { 2 } = 1 \text {, }$ is

A) $\frac { \pi } { 3 }$
B) $\frac { \pi } { 4 }$
C) $- \frac { \pi } { 4 }$
D) $- \frac { \pi } { 3 }$
E) $- \frac { \pi } { 2 }$

Question

Using Green's Theorem, the line integral $\int $C $x ^ { 2 } y d x - y ^ { 2 } x d y$ where C is the circle $x ^ { 2 } + y ^ { 2 } = 1 \text {, }$ is

A) $\frac { \pi } { 3 }$
B) $\frac { \pi } { 4 }$
C) $- \frac { \pi } { 4 }$
D) $- \frac { \pi } { 3 }$
E) $- \frac { \pi } { 2 }$

Quizplus · Accepted Answer

Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. For the given integral, $ P = x^2 y $ and $ Q = -y^2 x $. The partial derivatives are $\frac{\partial Q}{\partial x} = -y^2$ and $\frac{\partial P}{\partial y} = x^2$. The double integral over the region D is $\int \int_D (x^2 + y^2) \, dA$. Since $x^2 + y^2 = 1$ for the circle, the integral becomes $-\int \int_D 1 \, dA = -\pi$, giving the result $-\frac{\pi}{2}$ after considering the area of the circle.

Using Green's Theorem, the Line Integral ∫\int ∫ C x2ydx−y2xdyx ^ { 2 } y d x - y ^ { 2 } x d yx2ydx−y2xdy

Using Green's Theorem, the Line Integral $\int$ C $x ^ { 2 } y d x - y ^ { 2 } x d y$