Answer
$\frac{72}{5}$
Work Step by Step
$\int _{c}( x^\frac{2}{3}+y^2)dx +(y^\frac{4}{3}-x^2)dy$
In order to use Green Thereom's:
$\int _{c}Pdx+Qdy = \int\int _{D}( \frac{dQ}{dx} - \frac{dP}{dy})DA$
$\frac{dQ}{dx}= -2x$
$\frac{dP}{dy}= 2y$
Now plug these values into the formula
$\int\int _{D} ( -2x-2y) dA$
Now we need to check for values. When we look at the graph, we see that it is going clockwise, meaning our orientation is negative. Before solving the integral, we need to incorporate a negative before the integral to represent the negative orientation.
$-\int\int _{D} ( -2x-2y) dA$
Now we can look for x and y values.
If DA=dxdy
For x: 0 to $y^{2}$
For y: 0 to 2
If DA=dydx
y: 0 to $\sqrt x$
x: 0 to 4
I will be doing dxdy
$-\int ^{2} _{0}\int ^{y^2} _{0} -( 2x+2y) dxdy$
$\int ^{2} _{0}\int ^{y^2} _{0} ( 2x+2y) dxdy$
$\int ^{2} _{0} |x^2+2xy|^{y^2} _{0}dx$
$\int ^{2} _{0} x^4+2y^3dx$
$|\frac{x^5}{5}+\frac{y^4}{2}|^{2} _{0}$
$[\frac{2^5}{5}+\frac{2^4}{2}]$
$[\frac{32}{5}+8]$
$[\frac{32}{5}+\frac{40}{5}]$
Answer: $\frac{72}{5}$
