What Should Be the Values for E and F in a Triple Integral of a Sphere?

In summary, the conversation is discussing how to write an iterated triple integral to represent the volume of a spherical section with a radius of 5 cm and a thickness of 2 cm. The participants are discussing the values of E and F, which represent the lower and upper limits of z, and are trying to determine why their answers are not working. One participant suggests flipping the solid upside down to view it as an upper portion of a sphere centered at the origin, while another participant points out that this would not affect the volume. Eventually, the correct E and F values are determined and the answer is confirmed as correct. There is also a hint given in the problem about reflecting the solid over the xy plane, but this is found to be
  • #1
-EquinoX-
564
1

Homework Statement



The figure below shows part of a spherical ball of radius 5 cm. Write an iterated triple integral which represents the volume of this region.

http://img19.imageshack.us/img19/9237/sphereu.th.jpg

http://img19.imageshack.us/img19/1699/inte.jpg

Homework Equations


The Attempt at a Solution



I already got A, B, C and D correctly. However when I try E and F to be -3 and -sqrt(25-x^2-y^2) respectively and vice versa, the answer is wrong.. why is this.. what should be E and F then?
 
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  • #2
Is -3 really the minimum value of z for that volume?:wink:
 
  • #3
I guess so... z = -3 is just a plane right... other than that I really don't know what it is
 
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  • #4
Sure, but the region your interested in lies below the plane z=-3, so z=-3 should be the maximum not the minimum...likewise, the region of interest lies above the spherical surface z=-sqrt(25-x^2-y^2), so that should be the minimum value of z.

Did you try E=-sqrt(25-x^2-y^2) and F=-3?
 
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  • #5
yes I did and as I said on my original posting "vice versa" and it doesn't work as well..
 
  • #6
-EquinoX- said:
yes I did and as I said on my original posting "vice versa" and it doesn't work as well..

That's odd; assuming that E and F are the lower and upper limits on z, it should work...how do you know that it doesn't work? Are you imputing these values into an online submission? Or are you just getting an incorrect final answer for the volume?
 
  • #7
yes, into an online submission
 
  • #8
and shouldn't it be -sqrt(25-x^2-y^2) instead of -sqrt(25-x^2-z^2)
 
  • #9
Hmmm, what were your A,B,C and D?

Are you using the convention that [tex]\int_A^B\int_C^D\int_E^F dzdydx[/tex] means first integrate over z from E to F, then over y from C to D, and finally over x from A to B? Or are you using some other convention?
 
  • #10
A = -4
B = 4
C = -sqrt(16-x^2)
D = sqrt(16-x^2)
 
  • #11
-EquinoX- said:
and shouldn't it be -sqrt(25-x^2-x^2) instead of -sqrt(25-x^2-z^2)

Sorry, I meant to write -Sqrt(25-x^2-y^2)
 
  • #12
-EquinoX- said:
A = -4
B = 4
C = -sqrt(16-x^2)
D = sqrt(16-x^2)

That looks fine; you are either making a mistake/typo when typing in your E and F values, or the online submission program has a bug
 
  • #13
hmm...that's what I thought as well.. but are you sure that E=-sqrt(25-x^2-y^2) and F=-3?
 
  • #14
-EquinoX- said:
hmm...that's what I thought as well.. but are you sure that E=-sqrt(25-x^2-y^2) and F=-3?

Yes, why wouldn't that be correct? (Unless the volume is supposed to also include the cone that subtends that section of the sphere)
 
  • #15
Just making sure, as that's what I had in mind as well.. I'll have to ask my instructors about this then.. thanks for confirming
 
  • #16
Hmm.. there's a hint to this question, not sure if it helps

Hint: If you flip this solid upside down, you can view it as an upper portion of a sphere that is centered at the origin.
 
  • #17
-EquinoX- said:
Hmm.. there's a hint to this question, not sure if it helps

Hint: If you flip this solid upside down, you can view it as an upper portion of a sphere that is centered at the origin.

Hmmm... maybe they are expecting you to do that (even though it is clearly unnecessary); if so, then z would go from +3 to +Sqrt(25-x^2-y^2), so you could try those as your E and F values
 
  • #18
well..it makes no sense at all though... I only have one more chance to get this right
 
  • #19
Why doesn't it make any sense? The volume of the spherical section is the same no matter where it is right?...If the problem suggests that you reflect it over the xy plane, then you might as well...
 
  • #20
gabbagabbahey said:
Why doesn't it make any sense? The volume of the spherical section is the same no matter where it is right?...If the problem suggests that you reflect it over the xy plane, then you might as well...

yeah, I just realized that the volume is the same.. thanks for pointing that out.. now the answer is correct
 
  • #21
By writing "[itex]x^2+ y^2= 25[/itex] you are assuming this sphere is centered at the origin. That means that, since the sphere has radius 5 and the thickness of this portion is 3, the upper plane is NOT at x= -3. Hint: what is 5- 3?
 
  • #22
HallsofIvy said:
By writing "[itex]x^2+ y^2= 25[/itex] you are assuming this sphere is centered at the origin. That means that, since the sphere has radius 5 and the thickness of this portion is 3, the upper plane is NOT at x= -3. Hint: what is 5- 3?

Huh? the thickness of the portion is 2, not 3...
 
  • #23
what is the final answer about E and F?
 
  • #24
We don't provide "final answer" here. Read the PF Rules.

Zz.
 

1. What is a triple integral of a sphere?

A triple integral of a sphere is a mathematical concept used in calculus to calculate the volume of a three-dimensional spherical object. It involves integrating a function over a spherical region of space.

2. How is a triple integral of a sphere calculated?

To calculate a triple integral of a sphere, the spherical coordinates system is typically used. This involves converting the integrand and limits of integration from Cartesian coordinates to spherical coordinates, and then integrating over the three variables (radius, inclination angle, and azimuth angle) to find the volume.

3. What is the significance of a triple integral of a sphere in science?

The triple integral of a sphere has applications in a variety of scientific fields, including physics, engineering, and astronomy. It allows for the calculation of volumes and masses of spherical objects, as well as the calculation of moments of inertia and other physical properties.

4. What are the limits of integration for a triple integral of a sphere?

The limits of integration for a triple integral of a sphere depend on the specific problem being solved. Generally, the radius is integrated from 0 to the radius of the sphere, the inclination angle is integrated from 0 to π, and the azimuth angle is integrated from 0 to 2π.

5. Are there any practical examples of using a triple integral of a sphere?

Yes, there are many practical examples of using a triple integral of a sphere. For example, it can be used to calculate the volume of a water tank, the mass of a planet, or the moment of inertia of a spinning sphere. It can also be used in the calculation of electric fields and gravitational forces for spherical distributions.

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