19. use greens theorem to evaluate $\\int_{c}(e^{x}+y^{2})dx+(e^{y}+x^{2})dy$ where $c$ is the positively…

19. use greens theorem to evaluate $\\int_{c}(e^{x}+y^{2})dx+(e^{y}+x^{2})dy$ where $c$ is the positively oriented boundary of the region in the first quadrant bounded by $y = x^{2}$ and $y = 4$.

19. use greens theorem to evaluate $\\int_{c}(e^{x}+y^{2})dx+(e^{y}+x^{2})dy$ where $c$ is the positively oriented boundary of the region in the first quadrant bounded by $y = x^{2}$ and $y = 4$.

Answer

Explanation:

Step1: Recall Green's Theorem

Green's Theorem states that (\int_{C}Pdx + Qdy=\iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA), where (P = e^{x}+y^{2}), (Q = e^{y}+x^{2}). First, find (\frac{\partial Q}{\partial x}) and (\frac{\partial P}{\partial y}). (\frac{\partial Q}{\partial x}=2x), (\frac{\partial P}{\partial y}=2y). Then (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2x - 2y).

Step2: Determine the limits of integration

The curves (y = x^{2}) and (y = 4) intersect when (x^{2}=4), so (x = 2) (since we are in the first - quadrant, (x\geq0)). The region (D) is described as (0\leq x\leq2) and (x^{2}\leq y\leq4).

Step3: Set up the double - integral

(\iint_{D}(2x - 2y)dA=\int_{0}^{2}\int_{x^{2}}^{4}(2x - 2y)dydx). Integrate with respect to (y) first: (\int_{0}^{2}\left[(2xy-y^{2})\big|{y = x^{2}}^{y = 4}\right]dx=\int{0}^{2}\left[(2x\times4 - 4^{2})-(2x\times x^{2}-(x^{2})^{2})\right]dx). (=\int_{0}^{2}(8x - 16-2x^{3}+x^{4})dx).

Step4: Integrate with respect to (x)

(\int_{0}^{2}(8x - 16-2x^{3}+x^{4})dx=\left(4x^{2}-16x-\frac{1}{2}x^{4}+\frac{1}{5}x^{5}\right)\big|_{0}^{2}). (=4\times2^{2}-16\times2-\frac{1}{2}\times2^{4}+\frac{1}{5}\times2^{5}). (=16-32 - 8+\frac{32}{5}). (=(16-32 - 8)+\frac{32}{5}). (=-24+\frac{32}{5}=\frac{-120 + 32}{5}=\frac{-88}{5}).

Answer:

(-\frac{88}{5})