Solving an Integral Equation on a Curve C

[iRaid]

Homework Statement

Let the curve C be given by $\vec{r}(t)=3t^{2}\hat{\imath}-\sqrt{t}\hat{\jmath}$ between $0 \leq t \leq 4$. Calculate $\int_{C} xy^{2}dx+(x+y)dy$.

Homework Equations

The Attempt at a Solution

First find the derivative of r:
$$\vec{r}'(t)=6t\hat{\imath}-\frac{1}{2\sqrt{t}}\hat{\jmath}$$
$$\int_{C} xy^{2}dx+(x+y)dy=\int_{0}^{4} xy^{2}\hat{\imath}+(x+y)\hat{\jmath} \cdot (x'\hat{\imath}+y'\hat{\jmath})dt$$
$$\int_{0}^{4} (3t^{2})(\sqrt{t})^{2}\hat{\imath}+(3t^{2}-\sqrt{t})\hat{\jmath} \cdot (6t\hat{\imath}-\frac{1}{2\sqrt{t}}\hat{\jmath})dt$$
$$\int_{0}^{4}(3t^{3})(6t)+(3t^{2}-\sqrt{t})(\frac{-1}{2\sqrt{t}})dt=\int_{0}^{4} 18t^{4}+\frac{3t^{2}}{2\sqrt{t}}+\frac{1}{2}dt$$
$$=\frac{18t^{5}}{5}+\frac{\frac{3}{2}t^{\frac{5}{2}}}{\frac{5}{2}}+ \frac{t}{2}$$
Evaluate from 0 to 4.
=18538/5

I feel like I did something wrong...

 

[xiavatar]

Your answer is correct.

I am assuming it is just a typo, but be careful with your notation. Your missing some parentheses.

$$\int_{0}^{4} ((3t^{2})(\sqrt{t})^{2}\hat{\imath}+(3t^{2}-\sqrt{t})\hat{\jmath}) \cdot (6t\hat{\imath}-\frac{1}{2\sqrt{t}}\hat{\jmath})dt$$