Notes: Find the volume above the cone `z=sqrt(x^2+y^2)` and below the sphere `x^2+y^2+z^2=1`

Wednesday, 16 December 2015

Find the volume above the cone $\displaystyle{z}=\sqrt{{{{x}}^{{2}}+{{y}}^{{2}}}}$ and below the sphere $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}+{{z}}^{{2}}={1}$

The volume between two surfaces can be calculated by

$\displaystyle{V}=\int\int_{{D}}{\left({z}_{{{t}{o}{p}}}-{z}_{{{b}{o}{t}}}\right)}{d}{A}$

Where $\displaystyle{D}$ is the area contained by the boundary of the volume projected onto the xy-plane.

To find $\displaystyle{D}$ we need to solve when they two surfaces meet. Solve for $\displaystyle{z}$ on both equations and set them equal to each other.

We have $\displaystyle{z}=\sqrt{{{{x}}^{{2}}+{{y}}^{{2}}}}$

and

$\displaystyle{{z}}^{{2}}+{{x}}^{{2}}+{{y}}^{{2}}={1}$

$\displaystyle{z}=\pm\sqrt{{{1}-{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}}}$

Notice that the bottom half of the sphere $\displaystyle{z}=-\sqrt{{{1}-{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}}}$ is irrelevant here because it does not intersect with the cone. The following condition is true to find the curve of intersection.

$\displaystyle{z}={z}$

$\displaystyle\sqrt{{{{x}}^{{2}}+{{y}}^{{2}}}}=\sqrt{{{1}-{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}}}$

Switch to polar coordinates to simplify the problem.

$\displaystyle{r}=\sqrt{{{1}-{{r}}^{{2}}}}$

$\displaystyle{2}{{r}}^{{2}}={1}$

$\displaystyle{r}=\frac{{1}}{\sqrt{{{2}}}}$

The boundary of the volume is a circle of radius $\displaystyle\frac{{1}}{\sqrt{{{2}}}}$ . Then $\displaystyle{D}$ is

$\displaystyle{D}={\left\lbrace{\left({r},\theta\right)}{\mid}{0}\leq{r}=\lt\frac{{1}}{\sqrt{{{2}}}},{0}\leq\theta\leq{2}\pi\right\rbrace}$

Now preform the double integral.

$\displaystyle{V}=\int\int_{{D}}{\left[{\left(\sqrt{{{1}-{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}}}\right)}-{\left(\sqrt{{{{x}}^{{2}}+{{y}}^{{2}}}}\right)}\right]}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{\left({\left(\sqrt{{{1}-{{r}}^{{2}}}}\right)}-{r}\right)}{r}{d}{r}{d}{\left(\theta\right)}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{\left({{\left({1}-{{r}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}-{{r}}^{{2}}\right)}{d}{r}{d}{\left(\theta\right)}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{\left({r}{{\left({1}-{{r}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}-{{r}}^{{2}}\right)}{d}{r}{d}{\left(\theta\right)}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{\left({r}{{\left({1}-{{r}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right)}{d}{r}{d}{\left(\theta\right)}-{\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{{r}}^{{2}}{d}{r}{d}{\left(\theta\right)}$

On the first integral, let $\displaystyle{u}={1}-{{r}}^{{2}}$ $\displaystyle\to$ $\displaystyle{d}{u}=-{2}{r}{d}{r}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{r}{{\left({u}\right)}}^{{\frac{{1}}{{2}}}}{\left(\frac{{{d}{u}}}{{-{2}{r}}}\right)}{d}{\left(\theta\right)}-{\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{{r}}^{{2}}{d}{r}{d}{\left(\theta\right)}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{\left(-\frac{{1}}{{2}}\right)}{{\left({u}\right)}}^{{\frac{{1}}{{2}}}}{d}{u}{d}{\left(\theta\right)}-{\int_{{0}}^{{{2}\pi}}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{{r}}^{{2}}{d}{r}{d}{\left(\theta\right)}$

Integrate $\displaystyle{u}$ and substitute back in for $\displaystyle{r}$ .

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\left[{\left(-\frac{{1}}{{3}}\right)}{{\left({1}-{{r}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}-{\left(\frac{{1}}{{3}}\right)}{{r}}^{{3}}\right]}{{\mid}_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{d}{\left(\theta\right)}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\left[{\left(-\frac{{1}}{{3}}\right)}{{\left({1}-{{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}-{\left(\frac{{1}}{{3}}\right)}{{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}}^{{3}}+{\left(\frac{{1}}{{3}}\right)}\right]}{d}{\left(\theta\right)}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\left[{\left(-\frac{{1}}{{3}}\right)}{\left(\frac{{1}}{{{2}\sqrt{{{2}}}}}\right)}-{\left(\frac{{1}}{{{6}\sqrt{{{2}}}}}\right)}+{\left(\frac{{1}}{{3}}\right)}\right]}{d}{\left(\theta\right)}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}{\left[-\frac{{1}}{{{6}\sqrt{{{2}}}}}-\frac{{1}}{{{6}\sqrt{{{2}}}}}+\frac{{{2}\sqrt{{{2}}}}}{{{6}\sqrt{{{2}}}}}\right]}{d}{\left(\theta\right)}$

$\displaystyle{V}={\int_{{0}}^{{{2}\pi}}}\frac{{-{2}+{2}\sqrt{{{2}}}}}{{{6}\sqrt{{{2}}}}}{d}{\left(\theta\right)}$

$\displaystyle{V}={\left({2}\pi\right)}{\left(\frac{{-{2}+{2}\sqrt{{{2}}}}}{{{6}\sqrt{{{2}}}}}\right)}$

$\displaystyle{V}=\frac{{{2}\pi}}{{6}}\cdot{2}{\left(\frac{{-{1}+\sqrt{{{2}}}}}{\sqrt{{{2}}}}\right)}$

$\displaystyle{V}=\frac{{{2}\pi}}{{3}}\cdot{\left({1}-\frac{{1}}{\sqrt{{{2}}}}\right)}$

Notes

Wednesday, 16 December 2015

Find the volume above the cone $\displaystyle{z}=\sqrt{{{{x}}^{{2}}+{{y}}^{{2}}}}$ and below the sphere $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}+{{z}}^{{2}}={1}$

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Wednesday, 16 December 2015

Find the volume above the cone and below the sphere

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

Find the volume above the cone $\displaystyle{z}=\sqrt{{{{x}}^{{2}}+{{y}}^{{2}}}}$ and below the sphere $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}+{{z}}^{{2}}={1}$

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?