Finding bounds on triple integral?

In summary, the task is to integrate a function over a solid in the first octant bounded by two planes and contained in a sphere and cone. The equations for x, y, and z are given, as well as the formula for p. The hint suggests using horizontal slices of thickness dz to find the bounds for x, y, and z.
  • #1
beallio
2
0

Homework Statement



Integrate the function over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and and contained in a sphere centered at the origin with radius 20 and a cone opening upwards from the origin with top radius 16.



Homework Equations



x=psin(phi)cos(theta)
y=psin(phi)sin(theta)
z=pcos(phi)

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

The Attempt at a Solution


I don't understand how to get the x y and z bounds from the equations given.
 
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  • #2
Hi beallio! :smile:

Hint: horizontal slices of thickness dz, again, just like the other problem. :wink:
 

What is a triple integral?

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

Why do we need to find bounds on triple integrals?

Finding bounds on triple integrals allows us to define the limits of integration for each variable, which is necessary in order to accurately calculate the volume of a three-dimensional shape.

How do we find bounds on a triple integral?

The bounds for a triple integral are typically determined by the boundaries of the three-dimensional region being integrated over. This can be done by visualizing the shape or by using equations to define the boundaries.

What are some common methods for finding bounds on triple integrals?

Some common methods for finding bounds on triple integrals include using geometric shapes, setting up equations for the boundaries, and using symmetry to simplify the integral.

Why is it important to choose the correct bounds for a triple integral?

Choosing the correct bounds is important because it ensures that the integral is calculated accurately and provides the correct volume for the three-dimensional shape. Incorrect bounds can result in an incorrect volume calculation.

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