Evaluate the line integral $$\int_C \ x^2 dx+(x+y)dy \ $$
where $C$ is the path of the right triangle with vertices $(0,0), (4,0)$ and $(0,10)$ starts from the origin and goes to $(4,0)$ then to $(0,10)$ then back to the origin.
I did this problem but the answer is incorrect. First I looked at it from $(0,0)$ to $(4,0)$
THe magnitude of $r'(t)$ was $4$.
$r(t) = \langle 4t,0 \rangle$ for $0 \le t \le 1$.
Using this information I got $40/3$ as my answer after evaluating the integral.
Then I looked at the point from $(4,0)$ to $(0,10)$
$r(t)=\langle 4-4t,10t \rangle$ for $0 \le t \le 1$
The magnitude of $r'(t)$ is $2 \sqrt{29}$
Using this information I got $\frac{74}{3}\sqrt{29}$ as my answer after evaluating the integral.
Then I looked at the point from $(0,10)$ to $(0,0)$
$r(t) = \langle 0,-10t \rangle$ for $-1 \le t \le 0$
The magnitude of $r'(t)$ is $10$
Evaluating the integral I got the answer of $50$
Adding all three my final answer was $196.1673986$ but lon capa says this is incorrect. Can someone tell me where I am making a mistake?
Best Answer
You can simplify this considerably. The field is $$(x^2,x+y)=(x^2,y)+(0,x)$$ Note that the first component is conservative, so its line integral over a closed path is $0$. Therefore we are left with evaluating $$\int_C{xdy}$$ where $C$ is the right triangle. Now if you know Green's Theorem, then: By Green's Theorem, this will just be the area of the triangle. So the answer is $${1\over 2}\cdot4\cdot10=20$$
Otherwise, you should not have too much trouble computing that line integral along the triangle. Note that you can do one more simplification: it will be zero along the first edge, because the field $(0,x)$ is orthogonal to the line $y=0$.