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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

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

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

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

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

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

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

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