Integration in Polar - polar coordinate

In summary, the homework equation is f(x,y) = xy, where x ≥ 0 and y ≥ 0. The outer boundary is just a circle, and the upper limit for r is 2*sec(theta).
  • #1
DrunkApple
111
0

Homework Statement


f(x,y) = xy
x ≥ 0, y ≥ 0, [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ 4


Homework Equations





The Attempt at a Solution


[itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] rcosθ * rsinθ * r drdθ

I just wanted to check... is this right?? because I really don't think it is
 
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  • #2
DrunkApple said:

Homework Statement


f(x,y) = xy
x ≥ 0, y ≥ 0, [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ 4


Homework Equations





The Attempt at a Solution


[itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] rcosθ * rsinθ * r drdθ

I just wanted to check... is this right?? because I really don't think it is

Why are you putting the upper limit for r to be 2*sec(theta)? The outer boundary is just a circle, isn't it?
 
  • #3
Dick said:
Why are you putting the upper limit for r to be 2*sec(theta)? The outer boundary is just a circle, isn't it?
right! in terms of x, the bound should be from 0 to 2, but since x = 2 is rcos = 2, doesn't r becomes 2sec? or am I really really wrong and fault?
OH WAIT
is it just bounded by 0 to 2?

OH OH WAIT I JUST HAD AN EPIC EPIPHANY! THIS IS U SUBSTITUTION!
 
Last edited:
  • #4
DrunkApple said:
right! in terms of x, the bound should be from 0 to 2, but since x = 2 is rcos = 2, doesn't r becomes 2sec? or am I really really wrong and fault?
OH WAIT
is it just bounded by 0 to 2?

OH OH WAIT I JUST HAD AN EPIC EPIPHANY! THIS IS U SUBSTITUTION!

Uh, the boundary isn't x=2. It's x^2+y^2=4. u-substitution?? I think you are overcomplicating this.
 
  • #5
[itex]f^{pi/2}_{0}[/itex][itex]f^{2}_{0}[/itex] rcosθ * rsinθ * r drdθ
So it's not this?
 
  • #6
DrunkApple said:
[itex]f^{pi/2}_{0}[/itex][itex]f^{2}_{0}[/itex] rcosθ * rsinθ * r drdθ
So it's not this?

That's it.
 
  • #7
yes so I'll go head and calculate all this because there seems to be a problem and I don't know if it's calculation error or book error

[itex]f^{pi/2}_{0}[/itex][itex]f^{2secθ}_{0}[/itex] [itex]r^{3}[/itex]cosθ * rsinθ * drdθ

=[itex]f^{pi/2}_{0}[/itex]([itex]r^{4}[/itex]cosθsinθ)/4 * drdθ

=4[itex]f^{pi/2}_{0}[/itex]cosθsinθ dθ

After u substitution...

=4[itex]f^{1}_{0}[/itex]u du

=4?
 
  • #8
The integral from 0 to 1 of u*du isn't 1.
 
  • #9
oh oh wow... How could I miss that...
Thank you very much ;D
 

What is a polar coordinate system?

A polar coordinate system is a two-dimensional coordinate system used to locate points in a plane. It is based on a distance from the origin and an angle from a fixed reference line, typically the positive x-axis.

How is integration done in polar coordinates?

Integration in polar coordinates involves converting a double integral in Cartesian coordinates to a single integral in polar coordinates. This is done by converting the Cartesian coordinates to polar coordinates and adjusting the integrand accordingly.

What is the equation for converting Cartesian coordinates to polar coordinates?

The conversion from Cartesian coordinates (x, y) to polar coordinates (r, θ) is given by the equations r = √(x² + y²) and θ = arctan(y/x).

What are the limits of integration in polar coordinates?

The limits of integration in polar coordinates are determined by the shape being integrated over. The radial limit (r) is the distance from the origin to the outer boundary of the shape, and the angular limit (θ) is the angle between the positive x-axis and a line drawn from the origin to the outer boundary of the shape.

What are some common applications of integration in polar coordinates?

Integration in polar coordinates is commonly used in physics and engineering to solve problems related to circular motion, electric fields, and fluid dynamics. It is also used in mathematics to calculate areas and volumes of regions bounded by curves in polar coordinates.

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