Triple Integral Cartesian Coordinates

That would only cover the first quadrant. To cover the entire circle, you would need to integrate over all 4 quadrants, which could be done with the limits you gave and multiplying by 4.
  • #1
bytenel
2
0
Ok I have a quick question. I have this problem that is doable with polar coordinates and triple integrals but I was wondering if it would be possible to do this problem in the cartesian coordinate system (odd question I know...).

Homework Statement



A sprinkler distributes water in a circular pattern, supplying water to a depth of e^(-r) feet per hour at a distance of r feet from the sprinkler.

A. What is the total amount of water supplied per hour inside of a circle of radius 10?
2pi-2pie^(-10)
B. What is the total amount of water that goes throught the sprinkler per hour?
2pi

Homework Equations


Just integration techniques I guess. pi*R^2 is the equation for a circle area. x^2+y^2=100 is the equation in standard form for this circle.

The Attempt at a Solution



Here's where I get lost. In cartesian coordinates the bounds for the resulting double integral should be 0<=y<=sqrt(100-x^2) and 0<=x<=10, right? Then from there I take the double integral of the equation of the circle and...?

Any help is appreciated, thanks!

EDIT: Should I take the double integral of pi(x^2+y^2) with the bounds I have above? or is that wrong?
 
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  • #2
bytenel said:
Here's where I get lost. In cartesian coordinates the bounds for the resulting double integral should be 0<=y<=sqrt(100-x^2) and 0<=x<=10, right? Then from there I take the double integral of the equation of the circle and...?

Any help is appreciated, thanks!

EDIT: Should I take the double integral of pi(x^2+y^2) with the bounds I have above? or is that wrong?

Those bounds would be for a quarter of the circle. You could use those limits and multiply the answer by 4. But your integrand would be ##e^{-r}=e^{-\sqrt{x^2+y^2}}##. I don't think anyone would want to work it in rectangular coordinates. If you really must try it, you might have success with a trig substitution on the inside integral. Not sure, I haven't tried it.
 
Last edited:
  • #3
Wait, why would those limits only be for a quarter of the circle?
 
  • #4
bytenel said:
Wait, why would those limits only be for a quarter of the circle?

Because your limits have both x and y positive.
 

FAQ: Triple Integral Cartesian Coordinates

What is a triple integral in Cartesian coordinates?

A triple integral in Cartesian coordinates is a mathematical concept used to calculate the volume of a three-dimensional region. It involves integrating a function over a three-dimensional space using the three variables x, y, and z.

How is a triple integral in Cartesian coordinates different from a regular integral?

A regular integral involves integrating a function over a one-dimensional space, while a triple integral involves integrating a function over a three-dimensional space. This means that a triple integral requires three variables to be integrated, while a regular integral only requires one.

What is the purpose of using Cartesian coordinates in a triple integral?

Cartesian coordinates, also known as rectangular coordinates, are a standard way of representing points in three-dimensional space. Using Cartesian coordinates in a triple integral allows for a more efficient and accurate way to calculate the volume of a three-dimensional region.

What is the process for solving a triple integral in Cartesian coordinates?

The process for solving a triple integral in Cartesian coordinates involves dividing the three-dimensional region into smaller, simpler shapes (such as cubes or rectangular prisms), setting up the integral using the limits of integration for each variable, and then integrating the function over the three-dimensional space.

In what real-world applications are triple integrals in Cartesian coordinates used?

Triple integrals in Cartesian coordinates are used in many areas of science and engineering, such as calculating the mass of an object with varying density, determining the center of mass of a three-dimensional object, and solving problems in fluid mechanics and electromagnetics.

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