Question

Evaluate the line integral, where c is the given curve. ?c sin(xdx cos(ydy c consists of the top half of the circle x2 y2 = 9 from (3, 0 to (-3, 0 and the line segment from (-3, 0 to (-4, 4.

Answer 1

You can parameterize the curve $C$ and compute the integral along each component curve, or you can use the fact that $f(x,y)=(\sin x,\cos y)$ is the continuous gradient of a function $F(x,y)$ and observe that the line integral is path independent.

In other words, there is a function $F(x,y)$ such that $\nabla F(x,y)=f(x,y)$, so the integral along any curve $C$ from the points $\mathbf a$ to $\mathbf b$ is simply

$\displaystyle\int_Cf(x,y)\,\mathrm dS=\int_C(P(x,y)\,\mathrm dx+Q(x,y)\,\mathrm dy)=F(\mathbf b)-F(\mathbf a)$

You have $\nabla F(x,y)=(\sin x,\cos y)$, and so

$\dfrac{\partial F(x,y)}{\partial x}=\sin x\implies F(x,y)=-\cos x+g(y)$

while

$\dfrac{\partial F(x,y)}{\partial y}=\cos y=g'(y)\implies g(y)=\sin y+K$

where $K$ is an arbitrary constant. So we've found that

$F(x,y)=\sin y-\cos x+K$

which means the line integral has a value of

$F(-4,4)-F(3,0)=(\sin4-\cos 4)-(\sin0-\cos3)=\sin4-\cos4+\cos3$