Areas, volumes, arc length and physics

Find the area of the region enclosed by $y=x^{2}y\; =\; x^2$ and $y=x+2y\; =\; x\; +\; 2$.

Solution The curves meet where $x^2=x+2x^2\; =\; x\; +\; 2$, that is $(x-2)(x+1)=0(x\; -\; 2)(x\; +\; 1)\; =\; 0$, so $x=-1x\; =\; -1$ or $x=2x\; =\; 2$. At $x=0x\; =\; 0$, the line ($y=2y\; =\; 2$) is above the parabola ($y=0y\; =\; 0$), so the line is on top.

Use the disk method to find the volume of (a) a sphere of radius $rr$ (b) a cone of base radius $rr$ and height $hh$.

Solution

1. The sphere is the rotation of the semicircle $y=\sqrt{r^2-x^2}y\; =\; \backslash sqrt\{r^2\; -\; x^2\}$, $-r\le x\le r-r\; \backslash le\; x\; \backslash le\; r$: $$V=\pi {\int }_{-r}^r(r^2-x^2)dx=2\pi [r^{2}x-\frac{x^3}{3}{]}_0^r=2\pi (r^3-\frac{r^3}{3})=\frac{4}{3}\pi r^3\mathrm{.}V\; =\; \backslash pi\backslash int\_\{-r\}^\{r\}(r^2\; -\; x^2)\backslash ,dx\; =\; 2\backslash pi\backslash Big[r^2x\; -\; \backslash frac\{x^3\}\{3\}\backslash Big]\_0^\{r\}\; =\; 2\backslash pi\backslash left(r^3\; -\; \backslash frac\{r^3\}\{3\}\backslash right)\; =\; \backslash frac\{4\}\{3\}\backslash pi\; r^3\; .$$ (The integrand is even, so we doubled the integral from 0 to $rr$.)
2. The cone is the rotation of the line $y=\frac{r}{h}xy\; =\; \backslash frac\{r\}\{h\}x$, $0\le x\le h0\; \backslash le\; x\; \backslash le\; h$: $$V=\pi {\int }_0^h\frac{r^2}{h^2}{x}^{2}dx=\pi \frac{r^2}{h^2}\cdot \frac{h^3}{3}=\frac{1}{3}\pi r^{2}h\mathrm{.}V\; =\; \backslash pi\backslash int\_0^h\; \backslash frac\{r^2\}\{h^2\}x^2\backslash ,dx\; =\; \backslash pi\backslash frac\{r^2\}\{h^2\}\backslash cdot\backslash frac\{h^3\}\{3\}\; =\; \backslash frac\{1\}\{3\}\backslash pi\; r^2\; h\; .$$

(a) The region between $y=xy\; =\; x$ and $y=x^{2}y\; =\; x^2$ is rotated about the $xx$-axis. Find the volume. (b) The region between $y=2x-x^{2}y\; =\; 2x\; -\; x^2$ and the $xx$-axis is rotated about the $yy$-axis. Find the volume.

Solution

1. The curves meet at $x=0x\; =\; 0$ and $x=1x\; =\; 1$, and $x\ge x^{2}x\; \backslash ge\; x^2$ between them, so $R(x)=xR(x)\; =\; x$ and $r(x)=x^{2}r(x)\; =\; x^2$: $$V=\pi {\int }_0^1(x^2-x^4)dx=\pi (\frac{1}{3}-\frac{1}{5})=\frac{2\pi }{15}\approx \mathrm{0.419.}V\; =\; \backslash pi\backslash int\_0^1\; (x^2\; -\; x^4)\backslash ,dx\; =\; \backslash pi\backslash left(\backslash frac\{1\}\{3\}\; -\; \backslash frac\{1\}\{5\}\backslash right)\; =\; \backslash frac\{2\backslash pi\}\{15\}\; \backslash approx\; 0.419\; .$$
2. The region runs from $x=0x\; =\; 0$ to $x=2x\; =\; 2$. By shells: $$V=2\pi {\int }_0^{2}x(2x-x^2)dx=2\pi [\frac{2x^3}{3}-\frac{x^4}{4}{]}_0^2=2\pi (\frac{16}{3}-4)=\frac{8\pi }{3}\approx \mathrm{8.38.}V\; =\; 2\backslash pi\backslash int\_0^2\; x(2x\; -\; x^2)\backslash ,dx\; =\; 2\backslash pi\backslash Big[\backslash frac\{2x^3\}\{3\}\; -\; \backslash frac\{x^4\}\{4\}\backslash Big]\_0^2\; =\; 2\backslash pi\backslash left(\backslash frac\{16\}\{3\}\; -\; 4\backslash right)\; =\; \backslash frac\{8\backslash pi\}\{3\}\; \backslash approx\; 8.38\; .$$ Disks would need $xx$ in terms of $yy$, from $y=2x-x^{2}y\; =\; 2x\; -\; x^2$: two square-root branches. Shells avoid that.

Find the length of $y=\frac{2}{3}{x}^{\frac{3}{2}}y\; =\; \backslash frac\{2\}\{3\}x^\{\backslash frac\{3\}\{2\}\}$ from $x=0x\; =\; 0$ to $x=3x\; =\; 3$.

Solution $f^{\mathrm{\prime }}(x)=x^{\frac{1}{2}}f\text{'}(x)\; =\; x^\{\backslash frac\{1\}\{2\}\}$, so $1+[f^{\mathrm{\prime }}(x){]}^2=1+x1\; +\; [f\text{'}(x)]^2\; =\; 1\; +\; x$:

Check: the straight line from $(0,0)(0,\; 0)$ to $(3,2\sqrt{3})(3,\; 2\backslash sqrt\{3\})$ has length $\sqrt{9+12}\approx 4.58\backslash sqrt\{9\; +\; 12\}\; \backslash approx\; 4.58$. The curve must be a little longer, and it is.

(a) A particle moves with velocity $v(t)=t^2-4t+3v(t)\; =\; t^2\; -\; 4t\; +\; 3$ metres per second. Find its displacement and the distance it travels from $t=0t\; =\; 0$ to $t=3\mathrm{s}t\; =\; 3\backslash ,\backslash mathrm\{s\}$. (b) A spring has $k=200\mathrm{N}{\mathrm{m}}^{-1}k\; =\; 200\backslash ,\backslash mathrm\{N\backslash ,m^\{-1\}\}$. Find the work done in stretching it from $0.1\mathrm{m}0.1\backslash ,\backslash mathrm\{m\}$ to $0.3\mathrm{m}0.3\backslash ,\backslash mathrm\{m\}$ beyond its natural length.

Solution

1. Displacement: ${\int }_0^3(t^2-4t+3)dt=[\frac{t^3}{3}-2t^2+3t{]}_0^3=9-18+9=0\mathrm{m}\backslash int\_0^3\; (t^2\; -\; 4t\; +\; 3)\backslash ,dt\; =\; \backslash Big[\backslash frac\{t^3\}\{3\}\; -\; 2t^2\; +\; 3t\backslash Big]\_0^3\; =\; 9\; -\; 18\; +\; 9\; =\; 0\backslash ,\backslash mathrm\{m\}$: it ends where it started. But $v(t)=(t-1)(t-3)v(t)\; =\; (t\; -\; 1)(t\; -\; 3)$ is positive on $(0,1)(0,\; 1)$ and negative on $(1,3)(1,\; 3)$, so $$\text{distance}={\int }_0^{1}vdt-{\int }_1^{3}vdt=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\approx 2.67\mathrm{m}\mathrm{.}\backslash text\{distance\}\; =\; \backslash int\_0^1\; v\backslash ,dt\; -\; \backslash int\_1^3\; v\backslash ,dt\; =\; \backslash frac\{4\}\{3\}\; +\; \backslash frac\{4\}\{3\}\; =\; \backslash frac\{8\}\{3\}\; \backslash approx\; 2.67\backslash ,\backslash mathrm\{m\}.$$
2. $W={\int }_{0.1}^{0.3}200xdx=[100x^2{]}_{0.1}^{0.3}=100(0.09-0.01)=8\mathrm{J}W\; =\; \backslash int\_\{0.1\}^\{0.3\}\; 200x\backslash ,dx\; =\; \backslash Big[100x^2\backslash Big]\_\{0.1\}^\{0.3\}\; =\; 100(0.09\; -\; 0.01)\; =\; 8\backslash ,\backslash mathrm\{J\}$.