*Question 1* (a) Its volume is expressed by the following integrals: 1. $4 \int_{0}^{1} \int_{0}^{\frac{1}{2}\sqrt{1-x^{2}}} \int_{x^{2}+4y^{2}}^{1} \, dz \, dy \, dx$, 2. $4\int_{0}^{\frac{1}{2}} \int_{0}^\sqrt{ 1-4y^{2} } \int_{x^{2}+4y^{2}}^{1} \, dz \, dx \, dy$, 3. $4 \int_{0}^{1} \int_{0}^{\frac{1}{2}\sqrt{ z }} \int_{0}^{\sqrt{ z-4y^2 }} \, dx \, dy \, dz$. (b) Computing the first integral gives $\begin{align} V & = 4 \int_{0}^{1} \int_{0}^{\frac{1}{2}\sqrt{1-x^{2}}} \int_{x^{2}+4y^{2}}^{1} \, dz \, dy \, dx \\ & = 4 \int_{0}^{1} \int_{0}^{\frac{1}{2}\sqrt{1-x^{2}}} 1- x^{2} - 4y^{2} \, dy \, dx \\ &= 4 \int_{0}^{1} \left[ (1-x^{2})y -\frac{4}{3} y^{3} \right]^{\frac{1}{2}\sqrt{ 1-x^{2} }}_{0} \, dx \\ &= 4 \int \frac{1}{2} (1-x^{2})^{\frac{3}{2}} - \frac{1}{6} (1-x^{2})^{\frac{3}{2}} \, dx \\ & = \frac{4}{3} \int_{0}^{1} (1-x^{2})^{\frac{3}{2}} \, dx \\ &= \frac{4}{3} \int_{0}^{\frac{\pi}{2}} \cos^{4} \, du \\ &= \frac{4}{3} \int_{0}^{\frac{\pi}{2}} \left( \frac{\cos(2u) +1}{2} \right)^{2} \, du \\ &= \frac{1}{3} \int_{0}^{\frac{\pi}{2}} \cos^{2}(2u) +2\cos(2u) +1 \, du \\ & = \frac{1}{6} \int_{0}^{\frac{\pi}{2}} \cos 4u + 4\cos 2u + 3 \, du \\ & = \frac{1}{6} \left[ \frac{1}{4}\sin 4u + 2 \sin 2u + 3u \right]_{0}^{\frac{\pi}{2}} \\ &= \frac{1}{6} \times \frac{3\pi}{2} = \frac{\pi}{4} \end{align}$ *Question 2* (a) $\begin{align} M &=\int \int_{\Omega} \rho \, dA = \rho \int_{0}^{2\pi} \int_{0}^{1+\sin\theta} r \, dr \, d\theta \\ &= \rho \int_{0}^{2\pi} \left[ \frac{r^{2}}{2} \right]^{1+\sin\theta}_{0} \, dx \\ & = \frac{\rho}{2}\int_{0}^{2\pi} 1+\sin^{2} \theta + 2 \sin \theta \, d\theta \\ &= \frac{\rho}{2} \int _{0}^{2\pi} 1+ \frac{1}{2} - \cos 2 \theta + 2 \sin \theta \, d\theta \\ & = \frac{\rho}{2} \left[ \frac{3\theta}{2} -\frac{1}{4} \sin 2 \theta - 2\cos \theta\right]_{0}^{2\pi} \\ &= \frac{\rho}{2} [3\pi -2 +2 ] \\ & = \frac{3\pi\rho}{2} \end{align}$ (b) Note that the plate has reflectional symmetry in the $y$-axis since $r=1+\sin\theta = 1+\sin(\pi-\theta)$. Hence $\bar{x}=0$. Now $\begin{align} \int \int_{\Omega} \rho \,y \, dA &= \rho \int_{0}^{2\pi} \int_{0}^{1+\sin \theta } r^{2} \sin \theta \, dr \, d\theta \\ & = \rho \int_{0}^{2\pi} \left[ \frac{r^{3}}{3} \sin \theta \right]^{1+\sin \theta}_{0 } \, d\theta \\ & = \frac{\rho}{3} \int_{0}^{2\pi} (1+\sin \theta)^{3}(\sin \theta ) \, d\theta \\ &= \frac{\rho}{3} \int_{0}^{2\pi} \sin \theta + 3 \sin^{2}\theta + 3 \sin^{3} \theta + \sin^{4} \theta \, d\theta \tag{1} \end{align}$ For $n=0,1,2\dots$ define $I_{n}:=\int_{0}^{2\pi} \sin^{n} x \, dx$. For $n\geq 2$, integrating by parts gives $\begin{align} I_{n} &= \int_{0}^{2\pi} \sin^{n-1} x \sin x \, dx \\ & = [\sin^{n-1}x \cos x]_{0}^{2\pi} + \int_{0}^{2\pi} \cos x \cdot (n-1) \sin^{n-2} x \cos x \, dx \\ &= 0 + (n-1)\int_{0}^{2\pi} (1-\sin^{2}x)\sin^{n-2} x \, dx \\ & = (n-1)I_{n-2} + (n-1)I_{n } \\ \implies I_{n} &= \frac{n-1}{n} I_{n-2} \end{align}$ Now $I_{0}=\int_{0}^{2\pi } \, dx=2\pi$ and $I_{1}=\int_{0}^{2\pi }\sin x \, dx=0$. So $\int_{0}^{2\pi} \sin^{2} x \, dx =I_{2}= \frac{1}{2}I_{0}=\pi$ and $\int_{0}^{2\pi} \sin^{4} x \, dx = I_{4}= \frac{3}{4} I_{2} =\frac{3}{4}\pi$. Also $\int_{0}^{2\pi} \sin^{3} x \, dx = I_{3}= \frac{2}{3}I_{1}=0$. Hence substituting in $(1)$ gives: $\int \int_{\Omega} \rho \,y \, dA = \frac{\rho}{3} \left( 3\pi + \frac{3\pi}{4} \right) = \frac{5\pi \rho}{4}$So $\bar{y}= \frac{1}{M}\int \int_{\Omega} \rho \,y \, dA= \frac{2}{3\pi\rho} \times \frac{5\pi \rho}{4} =\frac{5}{6}$ and the plate has centroid $(0, 5/6)$. *Question 3 (a) We can evaluate the integral as follows $\begin{align} \text{Volume} &= \int_{0}^{2\pi} \, d\theta \int_{0}^{1} r(2-r-r^{2}) \, dr \\ &= 2\pi \left[ r^{2} -\frac{r^{3}}{3} -\frac{r^{4}}{4} \right]_{0}^{1} \\ &= 2\pi\left( 1-\frac{1}{3} -\frac{1}{4} \right) = \frac{5\pi}{6} \end{align}$ Below is a sketch of the solid: ![[Pasted image 20240224161304.png|400]] (b) Evaluation of the integral numerically using SciPy’s `tplquad` function: ![[Pasted image 20240224155813.png]]