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

Expert Answers
Charlie Hooper eNotes educator| Certified Educator

The volume between two surfaces can be calculated by

`V=int int_D (z_(t o p)-z_(b o t)) dA`

Where `D` is the area contained by the boundary of the volume projected onto the xy-plane.

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

We have `z=sqrt(x^2+y^2)`




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



Switch to polar coordinates to simplify the problem.




The boundary of the volume is a circle of radius `1/sqrt(2)` . Then `D` is

`D={(r,theta)|0lt= r =lt1/sqrt(2), 0lt= theta lt= 2pi}`

Now preform the double integral.

`V=int int_D [(sqrt(1-(x^2+y^2)))-(sqrt(x^2+y^2))] dxdy`

`V=int_0^(2pi) int_0^(1/sqrt(2)) ((sqrt(1-r^2))-r)r dr d(theta)`

`V=int_0^(2pi) int_0^(1/sqrt(2))((1-r^2)^(3/2)-r^2)dr d(theta)`

`V=int_0^(2pi) int_0^(1/sqrt(2))(r(1-r^2)^(1/2)-r^2)dr d(theta)`

`V=int_0^(2pi) int_0^(1/sqrt(2))(r(1-r^2)^(1/2))dr d(theta)-int_0^(2pi) int_0^(1/sqrt(2))r^2dr d(theta)`

On the first integral, let `u=1-r^2` `->` `du=-2r dr`

`V=int_0^(2pi) int_0^(1/sqrt(2)) r(u)^(1/2)((du)/(-2r)) d(theta)-int_0^(2pi) int_0^(1/sqrt(2))r^2dr d(theta)`

`V=int_0^(2pi) int_0^(1/sqrt(2)) (-1/2)(u)^(1/2) du d(theta)-int_0^(2pi) int_0^(1/sqrt(2))r^2dr d(theta)`

Integrate `u` and substitute back in for `r` .

`V=int_0^(2pi) [(-1/3)(1-r^2)^(3/2)-(1/3)r^3]|_0^(1/sqrt(2)) d(theta)`

`V=int_0^(2pi) [(-1/3)(1-(1/sqrt(2))^2)^(3/2)-(1/3)(1/sqrt(2))^3+(1/3)]d(theta)`

`V=int_0^(2pi) [(-1/3)(1/(2sqrt(2)))-(1/(6sqrt(2)))+(1/3)]d(theta)`

`V=int_0^(2pi) [-1/(6sqrt(2))-1/(6sqrt(2))+(2sqrt(2))/(6sqrt(2))] d(theta)`

`V=int_0^(2pi) (-2+2sqrt(2))/(6sqrt(2))d(theta)`




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