Homework 4

$$\gdef\diff#1#2{\frac{\mathrm{d}#1}{\mathrm{d}#2}} \gdef\dydx{\diff{y}{x}} \gdef\dd{\mathop{}\!\mathrm{d}} \gdef\nimplies{\;\;\;\not\nobreak\!\!\!\!\implies\;} \gdef\({\left(} \gdef\){\right)} \gdef\[{\left[} \gdef\]{\right]} \gdef\R{\mathbb{R}} \gdef\arraystretch{2.2em}$$

Mos Kullathon
921425216

1. In lecture, we found the volume of a donut by revolving the cricles $$(x-(R-r))^2 + y^2 = r^2$$ along the $y$-axis where $R>r$.

(a) Derive the integral formula for the volume of the donut. (It was covered in lass, but please rewrite it on your own)

First, we solve for $x$ to get an expression in terms of $y$.

$$(x-(R-r))^2 + y^2 = r^2 \\ (x-(R-r))^2 = r^2 - y^2 \\ x - (R-r) = \pm\sqrt{r^2-y^2} \\ x = (R-r) \pm\sqrt{r^2-y^2}$$

We can then split the expression for $x$ to produce two semicicles, where the positive square root component will produce the right side of the circle, and the negative component for the left side.

$$f_L(y) = (R-r) -\sqrt{r^2-y^2} \\ f_R(y) = (R-r) +\sqrt{r^2-y^2}$$

Revolving the right-hand component ($f_R(y)$) around the x-axis would give us the outside shape of the donut. Similarly, revolving the left-hand component ($f_L(y)$) around the x-axis would produce the inside portion (the hole) of the donut.

As such, the volume of the donut $V$ would be the volume produced by revolving the outside minus the inside.

$$\begin{align*} V &= \pi\int_{-r}^r f_R(y)^2 - f_L(y)^2 \dd y \\ &= \pi\int_{-r}^r\((R-r) + \sqrt{r^2-y^2}\)^2 - \((R-r) -\sqrt{r^2-y^2}\)^2 \dd y \end{align*}$$

Let $A = R-r$ and $B = \sqrt{r^2-y^2}$. Then,
$$\begin{align*} &\((R-r) + \sqrt{r^2-y^2}\)^2 - \((R-r) -\sqrt{r^2-y^2}\)^2 \\ &= (A+B)^2 - (A-B)^2 \\ &= (\cancel{A^2+B^2}+2AB) - (\cancel{A^2+B^2}-2AB) \\ &= 2AB - (-2AB) \\ &= 4AB \\ &= 4(R-r)\sqrt{r^2-y^2}. \end{align*}$$

And so, we have that:

$$\begin{align*} V &= \pi\int_{-r}^r\((R-r) + \sqrt{r^2-y^2}\)^2 - \((R-r) -\sqrt{r^2-y^2}\)^2 \dd y \\ &= \pi\int_{-r}^r 4(R-r)\sqrt{r^2-y^2} \dd y \\ &= 4\pi(R-r) \int_{-r}^r\sqrt{r^2-y^2}\dd y. \end{align*}$$

(b) Compute the integral and find the volume of the donut. (Substitute $y = r\sin\theta$)

With the integrand being in form $\sqrt{r^2-y^2}$, we can use trigonometic substitution.

Let $y = r\sin\theta$.

$$\begin{darray}{cc} y = r\sin\theta &\implies& \dd y = r\cos\theta\dd\theta \\ y = r\sin\theta = -r &\implies& \theta = \sin^{-1}(-1) = -\frac{\pi}{2} \\ y = r\sin\theta = r &\implies& \theta = \sin^{-1}(1) = \frac{\pi}{2} \end{darray} \\ \begin{align*} \therefore \int_{-r}^r \sqrt{r^2-y^2}\dd y &= \int_{-\pi/2}^{\pi/2} \sqrt{r^2-r^2\sin^2\theta}\cdot r\cos\theta\dd\theta \\ &= \int_{-\pi/2}^{\pi/2} \sqrt{r^2(1-\sin^2\theta)}\cdot r\cos\theta\dd\theta \\ &= \int_{-\pi/2}^{\pi/2} \sqrt{r^2\cos^2\theta}\cdot r\cos\theta\dd\theta \\ &= \int_{-\pi/2}^{\pi/2} r\cos\theta\cdot r\cos\theta\dd\theta \\ &= \int_{-\pi/2}^{\pi/2} r^2\cos^2\theta \dd\theta \\ &= r^2\int_{-\pi/2}^{\pi/2} \cos^2\theta\dd\theta \\ &= r^2\int_{-\pi/2}^{\pi/2} \frac{1+\cos2\theta}{2}\dd\theta \\ &= \frac{r^2}{2}\int_{-\pi/2}^{\pi/2} 1+\cos2\theta\dd\theta \\ &= \frac{r^2}{2}\[\theta+\frac{\sin2\theta}{2}\]_{-\pi/2}^{\pi/2} \\ &= \frac{r^2}{2}\(\frac{\pi}{2}+\cancel{\frac{\sin\pi}{2}} + \frac{\pi}{2}-\cancel{\frac{\sin(-\pi)}{2}}\) \\ &= \frac{\pi r^2}{2} \end{align*}$$

Finally, we substitute the result of this integral into our expression for volume.

$$\begin{align*} V &= 4\pi(R-r)\int_{-r}^r\sqrt{r^2-y^2}\dd y \\ &= 4\pi(R-r)\frac{\pi r^2}{2} \\ &= 2\pi^2r^2(R-r) \end{align*}$$

2. Find the volume of the solid of revolution by rotating the region bounded by $y = 2x^2, y = 8$ and the $y$-axis.

Question open to interpretation
Axis of revolution not specified. Assuming revolution around the y-axis.

$$y = 2x^2 \implies x = \pm\sqrt{\frac{y}{2}}$$

Since the region is bounded by the y-axis, we only consider the positive component.
$$x = \sqrt{\frac{y}{2}}$$

And at $x=0$, $y=0$. So, the integral for the volume $V$ is
$$\begin{align*} V &= \pi\int_0^8 \(\sqrt{\frac{y}{2}}\)^2\dd y \\ &= \pi \int_0^8 \frac{y}{2}\dd y \\ &= \pi \[\frac{y^2}{4}\]_0^8 \\ &= 16\pi. \end{align*}$$

Long Question 1

In Figure 1, the region $R$ is enclosed by the parabola $y = 4 − (x − 3)^2$ and the line segment $AC$ where $A, C$ are the points $(1, 0)$ and $(5, 0)$ respectively. $B$ is the vertec ofthe parabola.

(a) (i) Write down the coordinates of $B$.

$$\begin{darray}{cc} y = 4 - (x-3)^2 &\implies& \frac{\dd y}{\dd x} = -2(x - 3) \\ -2(x - 3) = 0 &\implies& x = 3 \\ x = 3 &\implies& y = 4-(3-3)^2 = 4 \end{darray} \\ \therefore B = (3, 4)$$

(ii) Write down the equation of the curve $AB$ and $BC$ as a function of $x = f(y)$.

$$y = 4-(x-3)^2 \\ y-4 = -(x-3)^2 \\ -y+4 = (x-3)^2 \\ \pm\sqrt{-y+4} = x - 3 \\ x = 3\pm\sqrt{-y+4} \\ \begin{align*} \therefore f_{AB}(y) &= 3 - \sqrt{4-y} \\ f_{BC}(y) &= 3 + \sqrt{4-y} \end{align*}$$

A jelly ring in the shape of solid of revolution of the region $R$ is revolved the $y$-axis as shown in Figure 2. The jelly ring contains two layers and we let $h$ be the height of the lower layer.

(b) (i) Show that the volume of the lower layer of the jelly ring is $$8\pi(8-(4-h)^{3/2})$$

Similar to question one, revolving the right-hand side ($f_{BC}(y)$) will produce the volume of the entire jelly. Revolving the left-hand side ($f_{AB}(y)$) will produce the volume of the hole.

As such, the volume of the lower portion of the jelly ring $V_\text{lower}$ would be the volume produced by revolving the outside minus the inside from $0$ to height $h$.

$$\begin{align*} V_\text{lower} &= \pi \int_0^h f_{BC}(y)^2-f_{AB}(y)^2 \dd y \\ &= \pi\int_0^h (3 + \sqrt{4-y})^2 - (3 - \sqrt{4-y})^2 \dd y \end{align*}$$

Let $A = 3$ and $B = \sqrt{4-y}$. Then,
$$\begin{align*} &(3 + \sqrt{4-y})^2 - (3 - \sqrt{4-y})^2 \\ &= (A+B)^2 - (A-B)^2 \\ &= 4AB \\ &= 12\sqrt{4-y}. \end{align*}$$

And so, we have that:
$$\begin{align*} V_\text{lower} &= \pi\int_0^h (3 + \sqrt{4-y})^2 - (3 - \sqrt{4-y})^2 \dd y \\ &= \pi\int_0^h 12\sqrt{4-y} \dd y \\ &= 12\pi\int_0^h \sqrt{4-y} \dd y \\ &= 12\pi\int_0^h (4-y)^{1/2} \dd y \\ &= 12\pi\[-\frac{2(4-y)^{3/2}}{3}\]_0^h \\ &= 12\pi \(-\frac{2(4-h)^{3/2}}{3} + \frac{2(4-0)^{3/2}}{3}\) \\ &= 12\pi \(-\frac{2(4-h)^{3/2}}{3} + \frac{16}{3}\) \\ &= 12\pi \(\frac{1}{3}\)(-2(4-h)^{3/2} + 16) \\ &= 12\pi \(\frac{1}{3}\)(2)(-(4-h)^{3/2} + 8) \\ &= 8\pi(-(4-h)^{3/2} + 8) \\ &= 8\pi(8-(4-h)^{3/2}) \end{align*}$$

which matches the provided expression for volume.

(ii) If the upper and the lower layers have the same volume, find the value of $h$. (You leave your answer of $h$ in three decimal places).

Since the integral found for the lower portion is valid for the entire jelly ring, we simply need to change its bound to reflect the upper portion.
$$\begin{align*} V_\text{upper} &= \pi\int_h^4 12\sqrt{4-y}\dd y \\ &= 12\pi\int_h^4 \sqrt{4-y}\dd y \\ &= 12\pi\[-\frac{2(4-y)^{3/2}}{3}\]_h^4 \\ &= 12\pi \(-\cancel{\frac{2(4-4)^{3/2}}{3}} + \frac{2(4-h)^{3/2}}{3}\) \\ &= \frac{24\pi(4-h)^{3/2}}{3} \\ &= 8\pi (4-h)^{3/2} \end{align*}$$

Since $V_\text{upper} = V_\text{lower}$, we can then solve for $h$.
$$\begin{align*} V_\text{upper} &= V_\text{lower} \\ \cancel{8\pi} (4-h)^{3/2} &= \cancel{8\pi}(8-(4-h)^{3/2}) \\ \sqrt{(4-h)^3} &= 8-\sqrt{(4-h)^3} \\ 2\sqrt{(4-h)^3} &= 8 \\ \sqrt{(4-h)^3} &= 4 \\ (4-h)^3 &= 16 \\ 4-h &= \sqrt[3]{16} \\ h &= -2\sqrt[3]{2} + 4 \\ &\approx 1.480 \end{align*}$$

(c) If milk is poured into the center of the jelly ring until it is fully filled. Find the volume of the milk required.

Revolving $f_{AB}(y)$ around the y-axis produces the volume of the hole. So, its volume is also the volume of the milk.

$$\begin{align*} V_\text{milk} &= \pi\int_0^4 f_{AB}(y)^2 \dd y \\ &= \pi\int_0^4 (3-\sqrt{4-y})^2 \dd y \\ &= \pi\int_0^4 (9+(4-y) - 6\sqrt{4-y}) \dd y \\ &= \pi\int_0^4 \(13 - y - 6(4-y)^{1/2}\) \dd y \\ &= \pi\[13y - \frac{y^2}{2} + \frac{12(4-y)^{3/2}}{3}\]_0^4 \\ &= \pi\[13y - \frac{y^2}{2} + 4(4-y)^{3/2}\]_0^4 \\ &= \pi\(13(4) - \frac{16}{2} - 4(4)^{3/2}\) \\ &= 12\pi \end{align*}$$