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Use Green's Theorem to evaluate the line integral:

$$\int_C2xy^{3}dx+4x^{2}y^{2}dy$$ and $C$ is a boundary of the triangular region in the first quadrant, enclosed by the x-axis, the line $x=1$, and the curve $y=x^{3}$.

This is my work:

$$P=2xy^{3}, Q=4x^{2}y^{2}$$ $$\int\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)=\int_0^1\int_0^{x^{3}}\left(8xy^{2}-6xy^{2}\right)dydx=\int_0^1\left[\frac{2xy^{3}}{3}\right]_0^{x^{3}}=\int_0^1\left(\frac{2x^{10}}{3}\right)dx=\int_0^1\left(\frac{2x^{11}}{33}\right)=\frac{2}{33}$$

I don't have a solutions manual so I just wanted to know if my answer is right.

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Your calculation is almost correct.. Your $P$ and $Q$ should be $2xy^3$ and $4x^2y^2$, respectively. There's also an algebra mistake that propagates through your answer. Have another go, if you're still struggling, I'll walk you through it with an edit to this answer.

Also, it's generally a good habit to confirm the prerequisites of the theorem are all true as well. Depending on the level/type of your course, your lecturer may be looking for this in your answer. (Or if you are self-studying, you should be looking for this.. :))

The following is Green's theorem, as taken from the wikipedia article on the subject:

"Let $C$ be a positively oriented, piecewise smooth, simple closed curve in a plane, and let $D$ be the region bounded by $C$. If $P$ and $Q$ are functions of $(x,y)$ defined on an open region containing $D$ and have continuous partial derivatives there, then $$\oint_C Pdx+Qdy = \iint_D\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} dxdy$$ where the path of integration along $C$ is counter-clockwise"

Is your $C$ a postively orientated, piecewise smooth, simple closed curve in the plane? Are $P$ and $Q$ defined on an open region containing $D$ and are their partial derivatives continuous there?

The answers may be 'obvious' but you need to show this.

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  • $\begingroup$ alright, I appreciate it! I can't believe I didn't notice that before. $\endgroup$ – juliodesa Dec 7 '15 at 1:21
  • $\begingroup$ Okay I changed it! I hope it is correct this time, because everything went very smooth. How would I show what C is? $\endgroup$ – juliodesa Dec 7 '15 at 1:30
  • $\begingroup$ Yeah, looks good. Though you are missing a few $dx$'s. Look at the definitions i.e 'simple closed curve' = non-intersecting closed curve. Is your $C$ a non-intersecting closed curve? Try to verify the rest. It would be probably best to make another question if you have problems. $\endgroup$ – PepperSausage Dec 7 '15 at 20:33

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