Converting to Polar Coordinates

In summary, the problem is to convert the given integral from rectangular form to polar form. The region of integration is a semi-circle from -π/2 to π/2, with a lower limit of integration of x=y. The integral can be written as ∫ from 0 to 3/√2 ∫ from y to √(9-y^2) of xydxdy. It is important to have a good understanding of the region of integration in order to solve this type of problem. A sketch of the region can be helpful in understanding the limits of integration.
  • #1
1question
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Homework Statement



Convert ∫ from 0 to 3/√2 ∫ from y to √(9-y^2) of xydxdy to polar form.


Homework Equations



x2+y2=r2

The Attempt at a Solution



I found the equation x2+y^2=9 from the upper range of the second integral. So r=3. Therefore r ranges from 0 to 3. The integrand is now (rcosθ)(rsinθ)rdrdθ.

I'm having trouble with the range of integration of the first integral (0 to 3/√2). From the other information, I think that the area I'm integrating is from -π/2 to π/2. This is wrong. Any hints? Thanks.
 
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  • #2
1question said:

Homework Statement



Convert ∫ from 0 to 3/√2 ∫ from y to √(9-y^2) of xydxdy to polar form.


Homework Equations



x2+y2=r2

The Attempt at a Solution



I found the equation x2+y^2=9 from the upper range of the second integral. So r=3. Therefore r ranges from 0 to 3. The integrand is now (rcosθ)(rsinθ)rdrdθ.

I'm having trouble with the range of integration of the first integral (0 to 3/√2). From the other information, I think that the area I'm integrating is from -π/2 to π/2. This is wrong. Any hints? Thanks.

It's essential in these kinds of problems to have a good understanding of what the region of integration looks like. Can you describe, in words, what this region looks like?
What does the lower limit of integration, y, represent?
 
  • #3
Oh. Is it the straight line x=y?

The region I've got so far is a semi-circle from -pi/2 to pi/2. Not really sure how to deal with the x=y.
 
Last edited:
  • #4
Here's your integral:
$$ \int_{y = 0}^{3/\sqrt{2}} \int_{x = y}^{\sqrt{9 - y^2}}~xy~dx~dy$$
If you click on the integral, you can see the LaTeX I wrote to produce it.
The inner integration goes across, from left to right, from the line to your half circle. The outer integration goes up from y = 0 to y = 3/√2.

You should draw a sketch of this region.
 

FAQ: Converting to Polar Coordinates

What does it mean to convert to polar coordinates?

Converting to polar coordinates is a way to represent a point in a two-dimensional plane using a distance from the origin and an angle from a reference line. It is an alternative to the more commonly used Cartesian coordinates system, which uses x and y coordinates to locate a point on a graph.

How do you convert from Cartesian coordinates to polar coordinates?

To convert from Cartesian coordinates (x, y) to polar coordinates (r, θ), you can use the following formulas: r = √(x^2 + y^2) and θ = arctan(y/x). Alternatively, you can also use trigonometric identities to calculate the values of r and θ.

What is the purpose of converting to polar coordinates?

Converting to polar coordinates can be useful for representing points in a circular or symmetrical pattern, such as in polar graphs or when working with complex numbers. It can also make certain calculations, such as finding the distance between two points, easier to solve.

Can you convert from polar coordinates back to Cartesian coordinates?

Yes, you can convert from polar coordinates to Cartesian coordinates using the following formulas: x = r cos(θ) and y = r sin(θ). These equations use trigonometric functions to calculate the x and y coordinates based on the distance (r) and angle (θ) values.

Are there any limitations to using polar coordinates?

While polar coordinates can be useful in certain situations, they may not be the most efficient or practical method of representing points in a two-dimensional plane. They are also limited in their ability to represent points in three-dimensional space. Additionally, some calculations, such as finding the slope of a line, are more straightforward in Cartesian coordinates.

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