Question

For the vector field G? =(yexy+4cos(4x+y))i? +(xexy+cos(4x+y))j? G?=(yexy+4cos(4x+y))i?+(xexy+cos(4x+y))j?, find the line integral of G? G? along the curve CC from the origin along the xx-axis to the point (4,0)(4,0) and then counterclockwise around the circumference of the circle x2+y2=16x2+y2=16 to the point (4/2?,4/2?)(4/2,4/2).

Answer 1

$\mathbf G(x,y)=(ye^{xy}+4\cos(4x+y))\,\mathbf i+(xe^{xy}+\cos(4x+y))\,\mathbf j$

We're computing the line integral

$\displaystyle\int_C\mathbf G\cdot\mathrm d\mathbf r$

It looks like the circular part of $C$ should be along the circle $x^2+y^2=16$ starting at (4,0) and terminating at $\left(\dfrac4{\sqrt2},\dfrac4{\sqrt2}\right)$.

Because integrating with respect to a parameterization seems like it would be a pain, let's check to see if $\mathbf G$ is a conservative vector field. For this to be the case, if $\mathbf G(x,y)=P(x,y)\,\mathbf i+Q(x,y)\,\mathbf j$, then $\mathbf G$ is conservative iff $\dfrac{\partial P(x,y)}{\partial y}=\dfrac{\partial Q(x,y)}{\partial x}$.

We have $P(x,y)=ye^{xy}+4\cos(4x+y)$ and $Q(x,y)=xe^{xy}+\cos(4x+y)$. The corresponding partial derivatives are

$\dfrac{\partial P(x,y)}{\partial y}=e^{xy}(1+xy)-4\sin(4x+y)$
$\dfrac{\partial Q(x,y)}{\partial x}=e^{xy}(1+xy)-4\sin(4x+y)$

and so the vector field is indeed conservative.

Now, we want to find a function $G(x,y)$ such that $\nabla G(x,y)=\mathbf G(x,y)=\left(\dfrac{\partial G(x,y)}{\partial x},\dfrac{\partial G(x,y)}{\partial y}\right)$. We have

$\dfrac{\partial G(x,y)}{\partial x}=ye^{xy}+4\cos(4x+y)$

Integrating with respect to $x$ yields

$\displaystyle\int\frac{\partial G(x,y)}{\partial x}\,\mathrm dx=\int(ye^{xy}+4\cos(4x+y))\,\mathrm dx$
$G(x,y)=e^{xy}+\sin(4x+y)+g(y)$

Differentiating with respect to $y$ gives

$\dfrac{\partial G(x,y)}{\partial y}=\dfrac{\partial}{\partial y}\left[e^{xy}+\sin(4x+y)+g(y)\right]$
$xe^{xy}+4\cos(4x+y)=xe^{xy}+\cos(4x+y)+\dfrac{\mathrm dg(y)}{\mathrm dy}$
$\implies \dfrac{\mathrm dg(y)}{\mathrm dy}=0$
$\implies g(y)=C$

and so

$G(x,y)=e^{xy}+\sin(4x+y)+C$

Because $\mathbf G(x,y)$ is conservative, and a potential function exists, the line integral is path-independent and the fundamental theorem of calculus of line integrals applies, so we can evaluate the line integral by evaluating the potential function at the endpoints. We end up with

$\displaystyle\int_C\mathbf G\cdot\mathrm d\mathbf r=G\left(\frac4{\sqrt2},\frac4{\sqrt2}\right)-G(0,0)=e^8-1+\sin(10\sqrt2)$