Did I solve this line integral correctly?

Could you please just check if I solved this integral correctly?

$$\int_\gamma ((\frac{2x}{x^2+y^2}+\cos x)dx+\frac{2(x^2+y^2+y)}{x^2+y^2}dy)$$

where $\gamma = \{x(t)=e^{2t}\cos t, y(t)=te^{3t^2}\}$

Firstly I reordered the integral like (1) + (2):

$$\int_\gamma (\frac{2x}{x^2+y^2}dx+\frac{2y}{x^2+y^2}dy)+\int_\gamma (\cos x dx + 2dy)$$

Then solved (2):

$$\int_0^1 (\cos(x(t))x'(t)+2y'(t))dt=[\sin(e^{2t}\cos t)+2te^{3t^2}]_0^1=$$ $$=\sin(e^2 \cos 1)+2e^3-\sin 1$$

Then I verified that (1) is a conservative field ($\int (Mdx+Ndy)$): $M_y=\frac{4xy}{(x^2+y^2)^2}=N_x$

and looked for the potential function $F(x,y)$:

$$F(x,y)=\int Mdx = \int \frac{2x}{x^2+y^2}=\ln(x^2+y^2)+c(y)$$ then found $c'(y)$ with $F_y=N$

$$\frac{2y}{x^2+y^2}+c'(y)=\frac{2y}{x^2+y^2}$$

so $c'(y) = 0$ and $c(y)$ = k and $F(x,y)=\ln(x^2+y^2)+k$.

By path independence I solved (2) as:

$$F(x(1),y(1))-F(x(0),y(0))=\ln(e^4\cos(1)^2)+e^6)=\ln(\cos(1)^2+e^2)+4$$

So in total (1)+(2) is:

$$\ln(\cos(1)^2+e^2)+4+\sin(e^2 \cos 1)+2e^3-\sin 1$$

It took me ages to type this in, so I hope it's a "good" question.

-
As often happens, "solve" isn't actually the right word. "Find" or "evaluate" would be appropriate. –  Michael Hardy Jan 22 '12 at 21:27
add comment

1 Answer

I was bored and checked your algebra, which looks right. Since I'm very prone to algebra misstakes myself, I checked you result numerically, and it indeed seems right: http://pastebin.com/wSinrBg3 (compare lines 9 and 14). [This is using the CAS "SymPy".]

PS: In all honesty I don't think this is a terribly good question, since what you ask for is quite boring (In the sense that checking someone elses tedious algebra is). I guess you got lucky ^^.

-
Indeed, I guess I got the 2 upvotes only because it's pretty to look at. Thanks for the help anyway! –  andrea Jan 22 '12 at 21:02
add comment