Integration/Double Integrals Advice Required

  • Thread starter Nima
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In summary, the conversation discusses how to integrate the function f(x, y) = Sqrt(x^2 + y^2) over a region in the x-y plane bounded by two circles. The suggested method is to use polar coordinates, as the region is circular. Alternatively, the region can be split into three parts and integrated using rectangular coordinates.
  • #1
Nima
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Hey, my Q is:

"Integrate f(x, y) = Sqrt(x^2 + y^2) over the region in the x-y plane bounded by the circles r = 1 and r = 4 in the upper half-plane".

Well, I firstly sketched out the region I get as my area in the x-y plane. I deduced that the ranges for x and y are:

0 <= x <= 4
Sqrt[1 - x^2] <= y <= Sqrt[16 - x^2]

1.) Is this right?
2.) How do I then calculate the integral of f(x, y) over this region? I know I'm doing a double integral but I don't see how I can separate my variables...

Thanks
 
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  • #2
That integral is screaming polar coordinates! f(x,y)=r and the region is circular. It'll be really easy in polar coordinates.
 
  • #3
There is no need to post every problem twice...
 
  • #4
Use the polar coordinates where ds=r dr d(theta) or if you want to use rectangular use these limits

x 1 -> 4
y sqrt(1-x^2) -> sqrt(16-x^2)
 
  • #5
If he wanted to use rectangular, he'd need to split it into 3 parts because when |x|>1, there is no inner circle anymore. So, as x goes from -4 => 1, 0 < y < sqrt(16-x^2), as x goes from -1 => 1, sqrt(1-x^2) < y < sqrt(16-x^2), and as x goes from 1 => 4, 0 < y < sqrt(16-x^2) again. Or he could do the whole half-disk of radius 4 and subtract the half-disc of radius one from it.
 
  • #6
yes you are right, but i just ment that he could understand that in this example we can multiply my previous result with 2 to get the answer.
 

1. What is integration and why is it important?

Integration is a mathematical concept that involves finding the area under a curve. It is important because it allows us to solve a wide range of problems in mathematics, science, and engineering.

2. What is the difference between a single integral and a double integral?

A single integral is used to find the area under a curve in one dimension, while a double integral is used to find the volume under a surface in two dimensions. In other words, a single integral deals with functions of one variable, while a double integral deals with functions of two variables.

3. How do I choose the limits of integration for a double integral?

The limits of integration for a double integral depend on the shape and bounds of the region being integrated. It is important to visualize the region and choose the appropriate limits based on the given function and boundaries.

4. Can I use the same integration technique for all functions?

No, there are various integration techniques, such as substitution, integration by parts, and trigonometric substitution, that are used for different types of functions. It is important to understand the properties of the function and choose the appropriate technique for integration.

5. What are some common applications of double integrals?

Double integrals have various applications in physics, engineering, and economics. They are used to calculate the volume of solid objects, find the center of mass of two-dimensional objects, and compute probabilities in statistics, among others.

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