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 Posted: Wed Nov 15, 2017 12:58 pm

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 20
How can I calculate the volume between the surfaces given below using double/triple integrals and polar coordinates?

$$\begin{Bmatrix} F_2(x, y) & = & 2-x^2-y^2 \\\\ F_1(x, y)& = & \sqrt{x^2+y^2} \end{Bmatrix}$$

Last edited by Math_Mod on Wed Nov 15, 2017 2:05 pm, edited 1 time in total.

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 Posted: Wed Nov 15, 2017 3:05 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
The surface $F_1=\big\{\big(x,y,\sqrt{x^2+y^2}\,\big)\;|\; (x,y)\in\mathbb{R}^2\big\}$ is a cone and the surface $F_2=\big\{\big(x,y,2-x^2-y^2\big)\;|\; (x,y)\in\mathbb{R}^2\big\}$ is a hyperboloid. These two surfaces intersect at the circle $C=\big\{\big(x,y,1\big)\;|\; x^2+y^2=1\big\}$. (see picture)

Attachment:

volhc.png [ 35.64 KiB | Viewed 342 times ] volhc.png [ 35.64 KiB | Viewed 342 times ]

So the solid in question is $$S=\Big\{\big(x,y,z\big)\in\mathbb{R}^3\;|\; -1\leqslant {x}\leqslant 1\,,\; -\sqrt{1-x^2}\leqslant{y}\leqslant\sqrt{1-x^2}\,,\;\sqrt{x^2+y^2}\leqslant {z}\leqslant 2-x^2-y^2 \Big\}$$ and the volume of $S$ is
\begin{align*}
\mathop{\iiint}\limits_{S}dV&=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{\sqrt{x^2+y^2}}^{2-x^2-y^2 }dz\,dy\,dx\\
&=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}2-x^2-y^2 -\sqrt{x^2+y^2}\,dy\,dx\\
&\stackrel{(*)}{=}\int_{0}^{1}\int_{0}^{2\pi}(2-r^2 -r)\,r\,d\varphi\,dr\\
&=\int_{0}^{1}2\pi\,(2r-r^3 -r^2)\,dr\\
&=2\pi\,\Big[r^2-\frac{r^4}{4}-\frac{r^3}{3}\Big]_{0}^{1}\\
&=2\pi\,\frac{5}{12}\\
&=\frac{5\pi}{6}\,.
\end{align*}

$(*)$ Change of coordinate system from cartesian to polar.

_________________
Grigorios Kostakos

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 Posted: Wed Nov 15, 2017 4:23 pm

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 20
Thank you, your help was invaluable!

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