Verify Green's Theorem for region bounded by the lines $x=2$, $y=0$, $y=2x$

Question

Verify Green's Theorem for the region D bounded by the lines $x=2$, $y=0$, $y=2x$ and the functions $f(x,y)=(2x^2)y$, $g(x,y)=2x^3$.

I have been trying this question for far too long and I can't seem to get it right. On one side I get $32$ and on the other I get $128$.

Answer 1

We split up $\partial D$ into $C_1, C_2, C_3$, where: \begin{align*} C_1:\qquad (f, g) &= (2(1 - t), 4(1 - t)) &\text{where } t \in [0, 1] \\ C_2:\qquad (f, g) &= (2t, 0) &\text{where } t \in [0, 1] \\ C_3:\qquad (f, g) &= (2, 4t) &\text{where } t \in [0, 1] \\ \end{align*} Note that: \begin{align*} \int_{C_1} f \, dx + g \, dy &= \int_0^1 2[2(1 - t)]^2 \cdot 4(1 - t) \cdot -2 \, dt + \int_0^1 2[2(1 - t)]^3 \cdot -4 \, dt \\ &= 2\int_0^1 -2^6(1 - t)^3 \, dt = 2^7 \int_1^0 u^3 \, du = -2^5 \\ \int_{C_2} f \, dx + g \, dy &= \int_0^1 2(2t)^2 \cdot 0 \cdot 2 \, dt + \int_0^1 2(2t)^3 \cdot 0 \, dt = 0 \\ \int_{C_3} f \, dx + g \, dy &= \int_0^1 2(2)^2 \cdot 4t \cdot 0 \, dt + \int_0^1 2(2)^3 \cdot 4 \, dt = 2^6 \\ \end{align*} Thus, we conclude that: $$ \int_{\partial D} f \, dx + g \, dy = -2^5 + 0 + 2^6 = 32 $$ which matches what you computed via Green's Theorem.