Bounds for triple integral

In summary, the problem involves integrating the function f(x,y,z)=–6x+2y over a solid in the first octant bounded by the planes x=0 and y=sqrt((277/123))x, contained in a sphere centered at the origin with radius 25 and a cone opening upwards from the origin with top radius 20. The bounds for integration are z from 0 to 25, phi from 0 to pi/2-arctan(3/4), and theta from .983002 to pi/2. The cone radius is an increasing function of z and intersects with the sphere at a radius of 20. The equations for the sphere and cone can be entered into a Java program to visualize
  • #1
Tom McCurdy
1,020
1
Integrate the function f(x,y,z)=–6x+2y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y=sqrt((277/123))x and contained in a sphere centered at the origin with radius 25 and a cone opening upwards from the origin with top radius 20.

I am having trouble getting the bounds
i got
0 - p -25
0 -phi -pi/2-arctan(3/4)
.983002-theta-pi/2
 
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  • #2
Tom McCurdy said:
Integrate the function f(x,y,z)=–6x+2y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y=sqrt((277/123))x and contained in a sphere centered at the origin with radius 25 and a cone opening upwards from the origin with top radius 20.

I am having trouble getting the bounds
i got
0 - p -25
0 -phi -pi/2-arctan(3/4)
.983002-theta-pi/2
The cone radius is an increasing function of z starting with zero radius at the origion. The square of the cone radius is proportional to (x^2 + y^2) where the proportionality constant determines the slope of the cone surface cutting any vertical plane.

The sphere and the cone intersect where the cone radius is 20. Write the equation of the sphere. What is z when a plane slice parallel to x-y through the sphere is a circle of radius of 20? What must the slope of the cone surace be so that it intersects the sphere at this same value of z with a radius of 20?

z ranges from 0 to 25, and for every z value there is a radius, either of the cone or of a cut of the sphere. For any value of z, the radius can be calculated, and r(z)^2 = x^2 + y^2. The minimum value of y is a function of x given by the boundary equation y=sqrt((277/123))x, and the maximum value is r(z). The minimum value of x is zero and the maximum value is r(z). It looks to me like you want to integrate in the order y, x, z with the z integral in two pieces, one where r(z) is determined by the cone and a second where r(z) is determined by the sphere.

Can you visualize the surface? If you are Java enabled you can get an idea if you go here

http://www.fedu.uec.ac.jp/~yanto/java/surface/

Enter these two expressions in the windows for z1 and z2 and check both boxes.

(3*sqrt(x^2+y^2)/4)*(1-1/(1+exp(10*sqrt(y^2-227*x/123))))

sqrt(25^2-x^2-y^2)*(1-1/(1+exp(10*sqrt(y^2-227*x/123))))

Click on Calculate. Change both 20 in "Divisions" to 100 and set the ranges from 0 to 25 for x, y, and z. Click Options and select Color Spectrum Mode. Click Options some more to turn on the tick marks. Click on rotate and get a view of the cone and sphere surfaces bounded by x = 0 and y=sqrt((277/123))x. You can rotate manually using the mouse pointing in the plot region with click and drag. You can use this to get a top down view. It may be tough to do manual rotation with 100 divisions, but you can cut it back to 50 or even 20 to do rotations and then increase it again. Click Calculate after every change. You can chop off the cone and sphere outside the radius of intersection by multiplying each z expression by (1-1/(1+exp(10*sqrt(y^2-410*x/123)))). Don't worry about the drips; it's only ice cream.
 

Related to Bounds for triple integral

1. What is a triple integral?

A triple integral is a type of mathematical integration that involves finding the volume under a three-dimensional function. It is similar to a double integral, but it adds another dimension to the integration process.

2. What are the bounds for a triple integral?

The bounds for a triple integral are the limits of integration for the three variables (x, y, and z) that are being integrated over. These bounds can be determined by looking at the limits of each variable in the given function.

3. How do you set up a triple integral?

To set up a triple integral, you first need to determine the bounds for each variable. Then, you need to determine the order of integration, which is the order in which you integrate the variables. This is typically done by starting with the innermost integral and working your way outwards.

4. What is the purpose of using bounds in a triple integral?

The bounds in a triple integral help define the region of space that is being integrated over. This allows you to find the volume under a three-dimensional function within a specific region, rather than the entire space.

5. Can the bounds for a triple integral be negative?

Yes, the bounds for a triple integral can be negative. This is because the limits of integration are determined by the function being integrated over, and the function may have negative values. However, the bounds must always be consistent with the order of integration and the orientation of the coordinate system being used.

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