*Question 1* Note that for $r = \frac{1}{4},$ $\begin{align} x(\theta) &= \frac{3}{4} \cos \theta + \frac{1}{4} \cos 3 \theta \\ y(\theta) &= \frac{3}{4} \sin \theta - \frac{1}{4} \sin 3 \theta \end{align} \quad \quad t\in [0,2\pi]$so $\begin{align} \frac{d}{d\theta} x (\theta) &= -\frac{3}{4} \sin \theta -\frac{3}{4} \sin 3\theta \\ \frac{d}{d\theta} y(\theta) &= \frac{3}{4} \cos \theta - \frac{3}{4} \cos 3\theta \end{align}$Let $D$ denote the region enclosed by the curve $C$ parametrised by $(x(\theta), y(\theta)).$ Then by Green's Theorem with $P=-y$ and $Q = x$ $\begin{align} \int \int _{D} 2 \, dx dy &= \int_{C} \left( x \frac{dy}{d\theta} - y \frac{dx}{d\theta} \right) \, d\theta \\ & = \int_{0}^{2\pi} \left(\frac{3}{4} \cos \theta + \frac{1}{4} \cos 3 \theta \right)\left(\frac{3}{4} \cos \theta - \frac{3}{4} \cos 3\theta\right) \\ & \quad + \left(\frac{3}{4} \sin \theta + \frac{3}{4} \sin 3\theta \right) \left(\frac{3}{4} \sin \theta - \frac{1}{4} \sin 3 \theta\right) \, d\theta \\ & = \frac{1}{16} \int_{0}^{2\pi} 9 \cos^{2} - 6 \cos\theta \cos 3 \theta - 3 \cos^{2} 3 \theta \\ & \quad \quad \quad \quad 9\sin^{2} \theta + 6 \sin\theta \sin 3\theta - 3\sin^{2} \theta \, d\theta \\ &= \frac{1}{16} \int_{0}^{2\pi} 6 - 6 \cos 4 \theta \, d \theta = \frac{1}{16} \times 12 \pi -0 \\ & = \frac{3}{4} \pi \\ \end{align}$so the area of $D$ is given by $\int \int_{D} \, dx \, dy = \frac{3}{8} \pi.$ *Question 2* (a) Firstly $ \underline{r}_{u} \times \underline{r}_{v} = \begin{array}{|ccc|} \underline{i} & \underline{j} & \underline{k} \\ \cos v & \sin v & -\sin v \\ -u\sin v & u \cos v & -u\cos v \end{array} = \begin{pmatrix} 0 \\ u \\ u \end{pmatrix}$so choose $+$ sign since $u$ is positive and we want $\hat{\underline{n}}$ upward-pointing. Also $\nabla \times \underline{F} = \begin{array}{|ccc|} \underline{i} & \underline{j} & \underline{k} \\ \partial_{x} & \partial_{y} & \partial_{z} \\ y^{2} & x &z \end{array} = \begin{pmatrix} 0 \\ 0 \\ 1-2y \end{pmatrix} $Therefore $\begin{align} \int\int_{S} \nabla \times \underline{F} \cdot \underline{\hat{n}} \, dS &= \int_{0}^{2\pi} \int_{0}^{1} \begin{pmatrix} 0 \\ 0 \\ 1-2u\sin v \end{pmatrix} \cdot \begin{pmatrix} 0 \\ u \\ u \end{pmatrix} \, du \, dv \\ & = \int_{0}^{2\pi} \int_{0}^{1} u - 2u^{2} \sin v \, du \, dv \\ & = \int_{0}^{2\pi} \left[ \frac{u^{2}}{2} -\frac{2}{3}u^{3} \sin v \right]_{0}^{1} \, dx \\ &= \int_{0}^{2\pi} \frac{1}{2} - \frac{2}{3} \sin v \, dv \\ & = \pi \end{align}$ (b) The boundary curve $C$ of the surface is parametrised by $\underline{r}(t) = (\cos t, \sin t, 2 - \sin t), \quad t\in [0,2\pi].$So by Stoke's theorem $\begin{align} \int\int_{S} \nabla \times \underline{F} \cdot \underline{\hat{n}} \, dS &= \oint_{C} \underline{F} \cdot \, d\underline{r} \\ & = \int_{0}^{2\pi} \begin{pmatrix} \sin^{2} t \\ \cos t \\ 2 - \sin t \end{pmatrix} \cdot \begin{pmatrix} -\sin t \\ \cos t \\ -\cos t \end{pmatrix} \, dt \\ & = \int_{0}^{2\pi} -\sin^{3} t + \cos^{2} t -2\cos t + \sin t\cos t \, dt \\ & = \pi \end{align}$as required. *Question 3* (a) See code and plot below. ![[Pasted image 20240314104147.png]] ![[Pasted image 20240314104221.png]] (b) Firstly $\underline{r}_{u} \times \underline{r}_{v} = \begin{array}{|ccc|} \underline{i} & \underline{j} & \underline{k} \\ -\sin u \cos v & -\sin u \sin v & \cos u \\ -(2+\cos u) \sin v & (2+\cos u) \cos v & 0 \end{array} = \begin{pmatrix} -\cos u\cos v (2+\cos u) \\ \vdots \\ \vdots \end{pmatrix}$ so choose $-$ sign since we want $\underline{\hat{n}}$ outward-pointing. Choose $\underline{F}(x,y,z) = (x,0,0)$ then $\nabla \cdot F = \begin{pmatrix} \partial_{x} \\ \partial_{y} \\ \partial _{z} \end{pmatrix} \cdot \begin{pmatrix} x \\ 0 \\ 0 \end{pmatrix} =1 $Therefore using Divergence theorem $\begin{align} \int \int \int_{V} \, dV &= \int \int_{S} \underline{F} \cdot \underline{\hat{n}} \, dS \\ & = \int_{0}^{2\pi} \int_{0}^{2\pi} \begin{pmatrix} (2+\cos u)\cos v \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} \cos u\cos v (2+\cos u) \\ \vdots \\ \vdots \end{pmatrix} \, du \, dv \\ &= \int_{0}^{2\pi} \int_{0}^{2\pi} \cos u \cos^{2}v (4+4\cos u + \cos^{2} u ) \, du \, dv \\ & = \int_{0}^{2\pi} \cos^{2} v \, dv \int_{0}^{2\pi} 4 \cos u + 4\cos^{2} u + \cos^{3} u \, du \\ & = \pi \times 4\pi = 4\pi^{2} \end{align} $