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Line Integral

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data
    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##.


    2. Relevant equations



    3. 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...
     
  2. jcsd
  3. May 5, 2014 #2
    Your answer is correct.

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

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