triple integration

**correct** · January 6th 2009, 02:41 AM

use triple integration to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4. its urgent pls

**mr fantastic** · January 6th 2009, 03:51 AM

Originally Posted by correct

use triple integration to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4. its urgent pls

A start would be to note that:

1. $x^2 + y^2 + z^2 = 4z \Rightarrow z^2 - 4z = -x^2 - y^2 \Rightarrow (z - 2)^2 - 4 = -x^2 - y^2 \Rightarrow (z - 2)^2 = 4 - x^2 - y^2$ .

2. The line of intersection is in the plane $4 = 4z \Rightarrow z = 1$ . Therefore the line of intersection is $x^2 + y^2 = 3$ .

**correct** · January 6th 2009, 04:37 AM

thank you. pls what should i do next

**Danny** · January 6th 2009, 04:40 AM

Originally Posted by correct

thank you. pls what should i do next

Using a double integral and as Mr. F the intersection of the two volumes is a circle think polar coordinates.

**correct** · January 6th 2009, 06:30 AM

i don't know how to proceed

**mr fantastic** · January 7th 2009, 06:03 AM

Originally Posted by correct

i don't know how to proceed

Do you know how to represent the volume as a triple integral? Before answering, please re-read the thread and look at what you've been told so far ....

**correct** · January 7th 2009, 10:49 AM

i know its triple integral dxdydz transforming this to polar coordinates it becomes r2sinthetadthetadpsi. if i use this, i don't know the limits

**mr fantastic** · January 7th 2009, 07:19 PM

Originally Posted by correct

i know its triple integral dxdydz transforming this to polar coordinates it becomes r2sinthetadthetadpsi. if i use this, i don't know the limits

to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4.

you should calculate the volume enclosed by the two spheres and then subtract this volume from the volume of the sphere $x^2+y^2+z^2=4z$ .

To get the enclosed volume:

First draw a simple diagram diagram. Looking at it side-one in the xz-plane (y = 0) it's not hard to see that the top surface is $z = \sqrt{4 - x^2 - y^2}$ and the bottom surface is $z = 2 - \sqrt{4 - x^2 - y^2}$ . This supplies the z-integration terminals.

The x and y integration terminals define the region in the xy-plane defined by the circle $x^2 + y^2 = 3$ .

So integrate the triple integral first with respect to z and then switch to polar coordinates.

**DeMath** · January 8th 2009, 12:27 AM

Originally Posted by correct

use triple integration to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4. its urgent pls

You can use the spherical coordinates, it's easy to do for your problem.

$\begin{gathered}\left\{ \begin{gathered}x = r\sin \theta \cos \varphi , \hfill \\y = r\sin \theta \sin \varphi , \hfill \\z = r\cos \theta . \hfill \\ \end{gathered} \right. \hfill \\dxdydz = {r^2}\sin \left( \theta \right)drd\theta d\varphi \hfill \\ \end{gathered}$

Then the volume is $V = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 3}} \right.\kern-\nulldelimiterspace} 3}} {\sin \left( \theta \right)} d\theta \int\limits_2^{4\cos \theta } {{r^2}dr}$ .

The answer must be $\frac{{22\pi }}{3}$ .

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