Find the line integral. Have I set this problem up correctly?

Question

F, the vector field, $ = [xy, x^{2}y^{2}]$

C is a quarter-circle from (2,0) to (0,2) with center (0,0).

So r(t) = the parametric equation of C $=[2cost(t), 2sin(t)$

$$r'(t) = [-2sin{t}, 2cos{t}]$$

$$F(r(t)) = [4sin(t)cos(t), 4cos^{2}{t} * 4sin^{2}{t}]$$ $$F(r(t)) = [4sin(t)cos(t), 16cos^{2}{t} * sin^{2}{t}]$$

Is this right so far? Then I setup the integral like this?

$$\int_C F(r(t) \cdot r'(t)$$ $$\int_C [4sin(t)cos(t), 16cos^{2}{t} * sin^{2}{t}] \cdot [-2sin{t}, 2cos{t}]$$

Is that right so far? Can someone help me out to the finish? My book says the asnwer is 8/5

Answer 1

You have $$ \int_C [4sin(t)cos(t), 16cos^{2}{t} * sin^{2}{t}] \cdot [-2sin{t}, 2cos{t}] $$ In the parameterized space, $0 \leq t \leq \frac{\pi}{2}$. Therefore, $$ =8(\int_{0}^{\frac{\pi}{2}} 4sin^2(t)cos^3(t)dt-\int_{0}^{\frac{\pi}{2}}sin^2(t)cos(t)dt) \\\\ =8(\int_{0}^{\frac{\pi}{2}} 4sin^2(t)(1-sin^2(t))cos(t)dt-\int_{0}^{\frac{\pi}{2}}sin^2(t)cos(t)dt) $$ Let $u=sin(t), du=-cos(t)dt$ $$ =8(4\int_{1}^{0} (u^2-u^4)du-\int_{1}^{0}u^2du) \\ =8(4(-\frac{1}{3}+\frac{1}{5})+\frac{1}{3}) \\ =8(-\frac{1}{5}) =-\frac{8}{5} $$ It seems i'm off by a minus sign but i'll leave that up to you!