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• Notes

Evaluate the line integral by two methods: (a) directly and (b) using Green's Theorem.
$\oint_{C} y^{2} d x+x^{2} y d y, C \text { is the rectangle with vertices }(0,0),(5,0),(5,4), \text { and }(0,4)$

$\oint_{C} y^{2} d x+x^{2} y d y \quad(0,0) \quad(5,0) \quad(5,4) \quad(0,4)$

(1) direct line Integral

IF $y=const \quad d y=0$
IF $x=const \quad d x=0$

$C_{1} : \int_{0}^{5} y^{2} d x=_{0}^{5}| x y^{2}=5 y^{2}=0$

$C_{2} : \int_{0}^{4}{x^{2}} y d y=_{0}^{4}| 25 * \frac{y^{2}}{2}==25 * \frac{16}{2}=200$

$C_{3}: \int_{5}^{0} y^{2} d x=(4)^{2}*-5=-80$

$C_{4}: \int_{4}^{0} x^{2} y d y=0$

$\oint_{c} y^{2} d x+x^{2} y d y=\int_{c_{1}}+\int_{c_{2}}+\int_{c_{3}}+\int_{c_{4}}$

$=0+200-80+0=120$

(b) Green Theorem

$p=y^{2} \quad , \quad Q=x^{2} y$

$\int_{D} \int \left(\frac{\partial}{\partial x} Q-\frac{\partial}{\partial y} p\right) d x d y$

$=\int_{0}^{4} \int_{0}^{5} 2 x y-2 y$

$=\int_{0}^{5} \frac{2 x y^{2}}{2}-\left.\frac{2 y^{2}}{2}\right|_{0} ^{4} d x$

$=\int_{0}^{5} f(x(16)-16) d x$

$\frac{5^{2}}{2} * 16-16 * 5=120$