I have to find volume using triple integration of region bounded by \$z=4 - sqrtx^2 +y^2\$ and \$z=sqrt x^2 +y^2\$. I have seen that they are cones intersecting at \$z=2\$, but my problem is that in \$x\$-\$y\$ plane shadow seems to be split up, which I cannot do. Please belp me with this. Thanks

The shaddow in the \$xy\$ plane is indeed a circle, of radius \$2\$, so in cartesian:

\$\$V = int_-2^2int_-sqrt4-x^2^sqrt4-x^2 int_sqrtx^2+y^2^4-sqrtx^2+y^2 1 dz, dy, dx\$\$

or in polar (easier):

\$\$V = int_0^2piint_0^2int_r^4-r 1 r, dz, dr, d heta\$\$

By symmetry above/below the plane \$z=2\$ you can calculate the volume of the lower or upper portion and multiply by \$2\$:

\$\$V = 2int_0^2piint_0^2int_r^2 1 r, dz, dr, d heta ext or V = 2int_0^2piint_0^2int_2^4-r 1 r, dz, dr, d heta\$\$

Geometrically this is the volume of two cones each with base area \$B=pi 2^2\$ and height \$h=2\$. SO the volume should come out to be \$\$V=2cdot dfrac13Bcdot h = dfrac16pi3\$\$

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edited Dec 24 "14 at 1:46
answered Dec 24 "14 at 1:41

David PDavid P
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The region does not split up. The shadow on the \$xy\$ plane is the circle where \$z=4 - sqrtx^2+y^2\$ and \$z=sqrtx^2+y^2\$ meet. This is when the radius is equal to \$2\$. Your inferior limit for \$z\$ is the lower cone and superior limit is the upper cone.

You are watching: Z=4-x^2-y^2

EDIT: You seem to be getting the bounded volume by the cones wrong. It is not what you drew with dashes. It is the volume between the cones, that is, for \$\$sqrtx^2+y^2 leq z leq 4 - sqrtx^2+y^2.\$\$ For the full algebra see David Peterson"s answer. :)

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edited Dec 24 "14 at 2:31
answered Dec 24 "14 at 1:36

Mark FantiniMark Fantini
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