18. use greens theorem to evaluate $\\int_{c}-y^{3} d x+x^{3} d y$ where $c$ is a circle given by…

Explanation:

Step1: Recall Green's Theorem

Green's Theorem states that (\oint_{C}Pdx + Qdy=\iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA). Here (P = - y^{3}), (Q=x^{3}).

Step2: Calculate partial - derivatives

First, find (\frac{\partial Q}{\partial x}) and (\frac{\partial P}{\partial y}). (\frac{\partial Q}{\partial x}=\frac{\partial(x^{3})}{\partial x}=3x^{2}), (\frac{\partial P}{\partial y}=\frac{\partial(-y^{3})}{\partial y}=- 3y^{2}). Then (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=3x^{2}+3y^{2}).

Step3: Convert to polar coordinates

Since the region (D) (the disk (x^{2}+y^{2}\leq4)) is a circle of radius (r = 2), use polar coordinates (x=r\cos\theta), (y = r\sin\theta), and (dA=rdrd\theta). Also (x^{2}+y^{2}=r^{2}). The double - integral becomes (\iint_{D}(3x^{2}+3y^{2})dA=\int_{0}^{2\pi}\int_{0}^{2}3r^{2}\cdot r drd\theta).

Step4: Evaluate the double - integral

First, integrate with respect to (r): (\int_{0}^{2\pi}d\theta\int_{0}^{2}3r^{3}dr). (\int_{0}^{2\pi}d\theta=\left[\theta\right]{0}^{2\pi}=2\pi). (\int{0}^{2}3r^{3}dr=3\times\frac{r^{4}}{4}\big|_{0}^{2}=3\times\frac{16}{4}=12).

Answer:

(24\pi)