Problem determining p in triple integral

In summary, the question asks for the center of mass of a solid bounded by a hemisphere and a plane where the density is directly proportional to the distance from the xy-plane. The integral is set up as m = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{1} kzp^{2}sin\theta dpd\phid\theta, but there seems to be an error as p should be equal to the radius of the sphere, which is 21, not 1. It is suggested that there may be a typo in the original problem, as z should not be negative in the given equation. Further clarification is needed to correctly solve the problem.
  • #1
gr3g1
71
0
The question states:

Find the center of mass of the solid that is bounded by the hemisphere z = sqrt(21 - x ^2 - y^2) and the plane z = 0 if the density at a point P is directly proportional to the distance from the xy-plane.

I know that the integral is setup :
[itex]
m = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{1} kzp^{2}sin\theta dpd\phid\theta
[/itex]

How ever I do not see how p = 1 for this equation...

i know x^2 + y^2 + z^2 = p^2

any help would be appreciated... for all my other problems i would isolate x^2 + y^2 + z^2 and determine my p, however, for this problem, that isn't working...

thanks a lot
 
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  • #2
gr3g1 said:
The question states:

Find the center of mass of the solid that is bounded by the hemisphere z = sqrt(21 - x ^2 - y^2) and the plane z = 0 if the density at a point P is directly proportional to the distance from the xy-plane.

I know that the integral is setup :
[itex]
m = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{1} kzp^{2}sin\theta dpd\phid\theta
[/itex]

How ever I do not see how p = 1 for this equation...
So "p" (actually the Greek letter "rho": [itex]\rho[/itex]) is the radius of the sphere? If [itex]z= \sqrt{21- x^2- y^2}[/itex] then squareing both sides gives [itex]z^2= 21- x^2- y^2[/itex] or [itex]x^2+ y^2+ z^2= 21[/itex], a sphere of radius 21, not 1. Of course in the orginal form z cannot be negative so that is the upper hemisphere. If what you have is correct then the integral you want should be
[tex]\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2}\int_{\rho= 0}^{21} k\rho cos(\phi) \rho^2 sin(\phi) d\rho d\phi d\theta [/tex]
Notice that I have made two corrections: in spherical coordinates, [itex]z= \rho cos(\phi)[/itex] and the differential involves [itex]sin^2(\phi)[/itex], not [itex]sin^2(\theta)[/itex]. But it is true that the radius of the sphere, as given, is 21, not 1.

I wonder if there isn't a typo here and the formula was intended to be [itex]z= \sqrt{1- x^2- y^2}[/itex].

i know x^2 + y^2 + z^2 = p^2

any help would be appreciated... for all my other problems i would isolate x^2 + y^2 + z^2 and determine my p, however, for this problem, that isn't working...

thanks a lot
 

What is the purpose of determining p in a triple integral?

The variable p in a triple integral represents the density of a three-dimensional object. By determining p, we can calculate the total mass or volume of the object.

How do you determine the value of p in a triple integral?

The value of p can be determined by analyzing the given problem and identifying the density function, which is typically represented by ρ. Once the density function is known, p can be calculated by integrating ρ over the given region of integration.

What are the limitations of using a triple integral to determine p?

One limitation is that it can only be used for objects with constant density. Additionally, accurately determining the limits of integration can be challenging for complex shapes.

Can p be negative in a triple integral?

No, p represents density which is always a positive value. If the calculated value of p is negative, it means an error has been made in the calculation.

How does the value of p affect the overall result of a triple integral?

The value of p directly affects the total mass or volume calculated by the triple integral. A higher value of p indicates a denser object, resulting in a larger mass or volume.

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