What Are the Limits of Integration for a Sphere and Cone Intersection?

[tix24]

Homework Statement

sketch the solid region contained within the sphere, x^2+y^2+z^2=16, and outside the cone, z=4-(x^2+y^2)^0.5.

b) clearly identifying the limits of integration, (using spherical coordinates) set up the iterated triple integral which would give the volume bounded by the above. Do not evaluate the integral.

c)using the theorem of pappus find the volume bounded by the above.

Homework Equations

The Attempt at a Solution

i have drawn the sketch of a sphere and the cone inside. they both have a circular trace on the xy plane of both radius 4.

im stuck on part b and c

my triple integral in spherical coordinates is the following (im hopping to get some help here because i don't know if it is correct or not)

∫dtheta ∫ sin(phi) dphi ∫ (rho)^ drho
my limits are the following theta from 0 to 2π
phi from 0 to π/2
rho from 4 to 4/(cos(phi)+sin(phi))

please help me on the limits of integration and the volume using theorem of pappus.

 

[haruspex]

All looks fine, except that you have rho decreasing over its range, which is likely to give you a negative result.

 

[tix24]

All looks fine, except that you have rho decreasing over its range, which is likely to give you a negative result.

im getting confused because the cone is only in the top half of the sphere, and in the bottom half of the sphere we have nothing. i don't know if the integral which i have set up is correct or not.

 

[haruspex]

im getting confused because the cone is only in the top half of the sphere, and in the bottom half of the sphere we have nothing. i don't know if the integral which i have set up is correct or not.

I'm not sure whether you are wondering if you should be including some volume from the lower half (answer, no), or whether you have mistakenly done so in your integral (again, no). You have phi from 0 to pi/2, so top half only.

 

[tix24]

I'm not sure whether you are wondering if you should be including some volume from the lower half (answer, no), or whether you have mistakenly done so in your integral (again, no). You have phi from 0 to pi/2, so top half only.

why shouldn't i include the lower half of the sphere?

 

[haruspex]

why shouldn't i include the lower half of the sphere?

Because that would all be inside the cone.

 

[tix24]

Because that would all be inside the cone.

fair enough, how do i use the theorem of pappus, can you aid me in this procedure? i know that the formula for volume using the theorem of pappus is as follows v=2pi(Ybar)A(D) but i am having trouble setting it up.

 

[haruspex]

fair enough, how do i use the theorem of pappus, can you aid me in this procedure? i know that the formula for volume using the theorem of pappus is as follows v=2pi(Ybar)A(D) but i am having trouble setting it up.

First step is to define clearly what all of the symbols mean in the formula.

 

[tix24]

First step is to define clearly what all of the symbols mean in the formula.

2pi just represents the constant in the equation, Ybar represents the distance from the axis which we are going to rotate the region at and A(D) is the area of the region. if i am going to find the volume of the bounded area then i have to find the area of this region. i think i can do this by finding the area of the cone and finding the area of the top half of the sphere ie half of the sphere. the subbtract the area of the sphere from the area of the cone. i don't know how to deal with (Ybar) though

 

[haruspex]

2pi just represents the constant in the equation, Ybar represents the distance from the axis which we are going to rotate the region at and A(D) is the area of the region. if i am going to find the volume of the bounded area then i have to find the area of this region. i think i can do this by finding the area of the cone and finding the area of the top half of the sphere ie half of the sphere. the subbtract the area of the sphere from the area of the cone. i don't know how to deal with (Ybar) though

What you are calling Ybar represents the distance of the area's centroid from the axis of rotation. However, the name "Ybar' suggests rotation about the x axis, so for the present problem it should be Xbar.
You can find the Xbar in much the same way as you find the area. Think of the area as having a moment about the axis. If Xbar is the distance to the centroid and A is the area then the moment is A Xbar. The moment of the area between sphere and cone is the difference of the moments of the corresponding quarter circle and triangle.
Alternatively, you can find Xbar directly by the usual integration: $\int xy.dx/\int ydx$

 

[tix24]

Because that would all be inside the cone.

did you mean to say it is outside of the cone? because the cone is only in the top half of the sphere, it leaves the sphere after the top half. (the radius exceeds that of a sphere.)

 

[haruspex]

the cone is only in the top half of the sphere

The definition of the cone admits z < 0, in which region the sphere is entirely inside the cone and therefore outside the region we're interested in.

 

[tix24]

The definition of the cone admits z < 0, in which region the sphere is entirely inside the cone and therefore outside the region we're interested in.

the equation provided fro the cone starts at (0,0,4) and expands downwards, the trace of the cone on the xy plane is a circle of radius 4. the sphere has a radius of 4. the cone is inside the sphere, and in the xy plane they both have the same radius. the cone is no longer inside the sphere after it has passed through the xy plane. becauseof this the bottom half of the sphere is empty... what is your reasoning for the sphere being inside the cone?

 

[haruspex]

the cone is no longer inside the sphere after it has passed through the xy plane.

Exactly. For z < 0, the interior of the sphere lies inside the cone. The only points in the entire 3-space which are inside the sphere but outside the cone have z > 0.