Evaluate line integral $\int_C{x^2}{y^2}dx + 4xy^3dy$ over a triangle

Question

Could someone help me with this problem? I tried it but kept getting different answers:

Evaluate $\int_C{x^2}{y^2}dx + 4xy^3dy$ where C is the positively oriented triangle with vertices at $(0,0), (1,3)$, and $(0,3)$

1. directly;
2. using Green's Theorem

Thanks in advance.

Answer 1

Using Stokes in 3D (because I am too lazy to remember Green in 2D) I get: $$ A(x,y,z) = (x^2 y^2, 4 x y^3, 0)^T $$

\begin{align} \int\limits_{C = \partial S} A \cdot dr &= \int\limits_S \mbox{rot } A \cdot dS \\ &= \int\limits_S(0, 0,4y^3-2x^2 y)^T \cdot dS \\ &= \int\limits_0^1 dx \int\limits_{3x}^3 dy \, (4y^3-2x^2y) \\ &= \int\limits_0^1 dx \, [y^4-x^2y^2]_{y=3x}^{y=3} \\ &= \int\limits_0^1 dx \, (81-9x^2-81x^4+9x^4) \\ &= [-72/5 x^5 + 81x -3x^3]_0^1 \\ &= -72/5 + 81 - 3 \\ &= 318/5 \\ &= 63.6 \end{align}

And the line integrals: \begin{align} I &= \int\limits_{C} A \cdot dr \\ &= \int\limits_{C_1} A \cdot dr + \int\limits_{C_2} A \cdot dr + \int\limits_{C_3} A \cdot dr \\ &= I_1 + I_2 + I_3 \end{align} with the first line $C_1$ along $r = (x, 3x, 0)^T$ with $dr = (1,3,0)^T dx$: \begin{align} I_1 &= \int\limits_{C_1} A \cdot dr \\ &= \int\limits_0^1 (x^2(3x)^2,4x(3x)^3,0)^T \cdot (1, 3, 0)^T dx \\ &= \int\limits_0^1 9x^4 + 324x^4 dx \\ &= [333/5 \, x^5]_0^1 \\ &= 333/5 \end{align} the second line $C_2$ backwards along $r = (x, 3, 0)^T$ with $dr=(1, 0, 0)^T dx$: \begin{align} I_2 &= \int\limits_{C_2} A \cdot dr \\ &= \int\limits_1^0 (x^2 3^2,4x 3^3,0)^T \cdot (1, 0, 0)^T dx \\ &= \int\limits_1^0 9 x^2 dx \\ &= [ 3 x^3]_1^0 \\ &= -3 \end{align} and the third line $C_3$ down along $r = (0, y, 0)^T$ with $dr = (0,1,0)^T dy$: \begin{align} I_3 &= \int\limits_{C_3} A \cdot dr \\ &= \int\limits_3^0 (0,0,0)^T \cdot (0, 1, 0)^T dy \\ &= \int\limits_3^0 0 dy \\ &= 0 \end{align} This gives $$ I = 333/5 - 3 + 0 = 318/5 $$ If the 3D embedding is too annoying, just drop the last components from the vectors.

Answer 2

If you share what you've tried, we'd be better able to help you by tailoring our input to your approach. With nothing to go on, I'm going to assume that you're probably having difficulty with the Green's Theorem part, not the standard evaluation.

Recall that Green's Theorem says: $$\int_CP\,dx+Q\,dy = \iint_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} dA$$

In this case, we have $P = x^2+y^2$ and $Q = 4xy^3$. Thus, our line integral results in:

\begin{align} \int_C x^2y^2dx + 4xy^3dy &=\int_0^3\int_0^{y/3}\left(4y^3\right)-\left(2x^2y\right)dx\,dy\\ &= \int_0^3\frac{4}{3}y^4-\frac{2}{81}y^4\;dy\\ &= \frac{106}{81}\int_0^3y^4\;dy\\ &= \frac{106}{3^4}\frac{3^5}{5}\\ &= \frac{318}{5} \end{align}