Changing cartesian integrals to equivalent polar integrals.

In summary, the problem states that you are trying to find the radius of a circle with center at (6,6), and radius r. You incorrectly use the upper limit of integration for r, which causes the region of integration to be too small. You end up solving for r using the bounds of integration that you determined, and the result is correct.
  • #1
soccerboy10
3
0

Homework Statement



6...y
∫...∫ x dx dy
0 ...0

The Attempt at a Solution


for reference: π=pi

6...y
∫...∫ x dx dy
0 ...0

y=6=x=6. rcosΘ=6 r=6/cosΘ
goes to

π/2 6secΘ
∫ ...∫... r cosΘ r dr dΘ
π/4...0

π/2.....r=6secΘ
∫ ... [((r^3)/3)cosΘ] dΘ
π/4......r=0

π/2
∫ ...72 sec^2 Θ dΘ
π/4

72tan(π/2) - 72tan(π/4)

last i checked, tan(π/2) was undefined. where did i undergo an error?
 
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  • #2
soccerboy10 said:

Homework Statement



6...y
∫...∫ x dx dy
0 ...0

The Attempt at a Solution


for reference: π=pi

6...y
∫...∫ x dx dy
0 ...0

y=6=x=6. rcosΘ=6 r=6/cosΘ
goes to

π/2 6secΘ
∫ ...∫... r cosΘ r dr dΘ
π/4...0
The upper limit of integration for r is wrong. The region over which integration is taking place is the triangle bounded by the y axis, the line y = x, and the line y = 6. Along this horizontal line, r = 6 cscΘ.

soccerboy10 said:
π/2.....r=6secΘ
∫ ... [((r^3)/3)cosΘ] dΘ
π/4......r=0

π/2
∫ ...72 sec^2 Θ dΘ
π/4

72tan(π/2) - 72tan(π/4)

last i checked, tan(π/2) was undefined. where did i undergo an error?
 
  • #3
how do you integrate csc(cubed) cos?
 
Last edited:
  • #4
Please take a look at what I said in my previous post. The integrand is NOT sec3(Θ)cos(Θ)
 
  • #5
my apologies been a long day studying. problem ended out fine.
 
  • #6
Not to dig up an old thread, but I have this same problem. I thought this would be better than making an entirely new post. I follow the problem, but am confused as to how the bounds of integration for r were found? I drew the picture, and have the triangle bounded by y=6 at the top, the y axis, and line y=x. I can easily get theta from pi/4 to pi/2, but I'm struggling with finding r. Any advice? Also, this is something I've had issues with on a lot of polar coordinate problems. So, if anyone has a more general explanation as to r, instead of just giving an answer for r's bounds in this problem, that would be greatly appreciated!
 

1. What is the purpose of changing cartesian integrals to polar integrals?

The purpose of changing cartesian integrals to polar integrals is to simplify the integration process and make it easier to solve certain types of problems, particularly those that involve circular or symmetric regions.

2. How do you convert a cartesian integral to a polar integral?

To convert a cartesian integral to a polar integral, you must first identify the equation of the curve in polar coordinates. Then, you can substitute the appropriate polar variables (r, θ) into the integral and use the appropriate conversion formulae to change the limits of integration.

3. What are the benefits of using polar coordinates for integration?

Polar coordinates are particularly useful for integrating over circular or symmetric regions, as they can simplify the integrand and make it easier to solve. They also allow for a wider range of integration techniques, such as using symmetry or the polar substitution method.

4. Are there any limitations or drawbacks to using polar coordinates for integration?

One limitation of using polar coordinates for integration is that they are not as intuitive as cartesian coordinates, and may require more practice to understand and use effectively. Additionally, not all regions or curves can be easily described or integrated using polar coordinates.

5. Can any cartesian integral be converted to an equivalent polar integral?

No, not all cartesian integrals can be converted to an equivalent polar integral. Some integrals may be more complicated or impossible to express in terms of polar coordinates, while others may not benefit from the conversion and may be easier to solve in cartesian form.

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