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

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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
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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
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Mark FantiniMark Fantini
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