Having trouble with double integrals in polar coordinates?

In summary, the problem asks for the lower bound for r in terms of cosine and sinine, and the iterated integral provides a few tips on how to do this.
  • #1
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I'm having trouble figuring out how to find what "r" is. I know r is the radius, but how do I go about finding it? Like what do I look for in a particular problem?
 
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  • #2
Maybe if you gave the equation you'd get something more definite. But for polar coordinates what i'd do personally is convert it to a set of parametric equations and integrate those separately with respect to the parameter.

R is usually given by a polar equation, I'm not quite sure what you're asking.
 
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  • #3
@gordonj005: I mean I'm confused as to what steps to take to find the radius, r.
 
  • #4
again it would be helpful if you gave the context. the radius is just the distance from the origin. where do the double integrals fit in?
 
  • #5
Okay here is a problem from my homework, and I'm stuck on finding the lower bound for r:

Graph r=1/(9cos(t)) for -∏/2≤t≤∏/2 and r=1. Then write an iterated integral in polar coordinates representing the area inside the curve r=1 and to the right of r=1/(9cost) . (Use t for theta in your work.)

I got the bounds for t -arccos(1/9) ≤t≤ arccos(1/9) and I have the upper bound for r to be 1, but I can't get the lower bound, any ideas?

Thank you
 
  • #6
sorry, just to clarify.. [itex] r = \frac{1}{9 \cos{t}}[/itex] for [itex]-\pi \le t \le \frac{\pi}{2}[/itex] and [itex] r = 1 [/itex] right?

For the lower bound on r, I believe it is [itex]\frac{-1}{9} [/itex], but I'm not entirely sure about that, maybe someone else can look at this.

In terms of the iterated integral here's a few tips:
[tex] x = r \cos{t} [/tex]
[tex] y = r \sin{t} [/tex]
[tex] dA = r dr dt [/tex]

and you're integrating a function of the form [itex] f(x, y) [/itex].
 
  • #7
would you mind? can you please ans this... please find the polar form of 1/4i also the 1/z? please
 

1. What are polar coordinates and how are they used in double integrals?

Polar coordinates are a coordinate system used to represent points in a two-dimensional space. They consist of a distance from the origin (known as the radius) and an angle from a reference line (known as the polar angle). In double integrals, polar coordinates are used to simplify the integration process for certain types of functions, particularly those with circular or symmetric shapes.

2. How do I convert a double integral from Cartesian coordinates to polar coordinates?

To convert a double integral from Cartesian coordinates to polar coordinates, you will need to use the following formulas:

x = r cosθ

y = r sinθ

dx dy = r dr dθ

After substituting these formulas into the original integral, you can then integrate with respect to r and θ instead of x and y.

3. What is the significance of the Jacobian in double integrals with polar coordinates?

The Jacobian is a mathematical concept used to calculate the change of variables in an integral. In double integrals with polar coordinates, the Jacobian is equal to r, which is why it is included in the conversion formula (dx dy = r dr dθ). The Jacobian essentially helps to account for the change in area when converting from Cartesian to polar coordinates.

4. How do I determine the limits of integration for a double integral in polar coordinates?

The limits of integration for a double integral in polar coordinates depend on the shape and boundaries of the region being integrated. To determine the limits, you will need to sketch the region and identify the points of intersection with the polar axis (r = 0) and the boundaries of the region. The polar angle limits will also need to be determined based on the symmetry of the region.

5. Are there any tips for solving double integrals in polar coordinates?

One helpful tip for solving double integrals in polar coordinates is to always sketch the region and identify any symmetries. This will help determine the limits of integration and make the integration process easier. It is also important to carefully substitute the conversion formulas and pay attention to the Jacobian factor. Practice and familiarity with polar coordinates will also make solving double integrals easier.

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