# Triple Integral problem (maybe using spherical coordiantes)

• batmankiller
In summary, the homework statement is that the function f(xyz)=-4x+6y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y=x*sqrt(22/3) and contained in a sphere centered at the origin with radius 13 and a cone opening upwards from the origin with top radius 5. The attempt at a solution is to change function 6y-4x into the terms of spherical coordinates. While trying to find the limits of the integral, the student is not sure if their drawing is correct.
batmankiller

## Homework Statement

Integrate the function f(xyz)=−4x+6y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y=x*sqrt(22/3) and contained in a sphere centered at the origin with radius 13 and a cone opening upwards from the origin with top radius 5.

## Homework Equations

x^2+y^2+z^2=p(row)
x=p(sinf(fie)*cost(theta)
y=p*sinfsint

## The Attempt at a Solution

So I changed function 6y-4x into the terms of spherical coordinates. While i attempted to find the limits of the integral, I'm not sure if my drawing is correct:

But I assumed it meant that the region had to be inside the cone such that it was in the first octet (since it's bounded by x=0 which is the yz plane --NOT SHOWN IN PICTURE), but I'm not sure how the ice cream part has to do with this...
Could someone just explain to me what the trace part means and also explain if my drawing is right, I have no idea what inside the circle and the cone means..

I would recommend cylindrical coordinates because there is more cylindrical symmetry than spherical symmetry. If you alter the coordinates so that the axis of the cone is along the z-axis then you have cone itself as given by $z= \sqrt{22/3}r$ and the sphere by r^2+ z^2= 169. To integrate over that, $\theta$ goes from 0 to $2\pi$, r goes from 0 to 5, and z goes from $\sqrt{22/3}r$ to $\sqrt{169- r^2}$. Of course, now the integrand is $f(x, y, z)= -4x+ 6z= -4rcos(\theta)+ 6z$ since we have swapped y and z.

Oh so you do suggest that I change the cone? altering it doesn't mess with the problem?
i still can't seem to integrate this =[

Last edited:

## 1. What is a triple integral problem?

A triple integral problem is a mathematical calculation that involves finding the volume of a three-dimensional region within a given function or surface. It is an extension of a single and double integral, which solve for area and volume respectively.

## 2. What are spherical coordinates?

Spherical coordinates are a type of coordinate system used to specify points in three-dimensional space. It uses two angles, θ and φ, and a distance, ρ, from the origin to describe a point in terms of its radial distance, inclination, and azimuth.

## 3. How do you convert from rectangular to spherical coordinates?

To convert from rectangular coordinates (x, y, z) to spherical coordinates (ρ, θ, φ), use the following equations: ρ = √(x² + y² + z²), θ = tan⁻¹(y/x), and φ = tan⁻¹(√(x² + y²)/z).

## 4. How is a triple integral problem solved using spherical coordinates?

A triple integral problem using spherical coordinates is solved by integrating the given function over the spherical region. The integrand is multiplied by ρ²sinφ and the limits of integration are determined by the shape of the region in spherical coordinates.

## 5. What are some applications of triple integrals in science and engineering?

Triple integrals have a wide range of applications in various fields. They are commonly used in physics to calculate the mass and center of mass of three-dimensional objects. In engineering, they are used to determine the moment of inertia of complex objects. They are also used in fluid dynamics to calculate the volume and flow rate of fluids. Additionally, they have applications in astronomy, electromagnetism, and quantum mechanics.

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