Did you try to plot the region of integration? How did you get those limits?IsaacStats said:Well,
0<=theta<=pi
0.5<=r<=1
CAF123 said:Did you try to plot the region of integration? How did you get those limits?
Those bounds describe a portion of the unit circle centered on the origin in the first quadrant.IsaacStats said:I got 0<=theta<=pi/2, 0<=r<=1.
CAF123 said:Those bounds describe a portion of the unit circle centered on the origin in the first quadrant.
Are you sure you are considering the right half of the curve x = ±√(y-y2)? For sure, this exists in both first and second quadrants, but you are only interested in the half x = -√(y-y2). Since √(y-y2) ≥ 0, what quadrant does x=-√(y-y2) lie?IsaacStats said:So, to get to the right conclusion do I add the area of the half of the circle centered at (0,0.5)?
CAF123 said:Are you sure you are considering the right half of the curve x = ±√(y-y2)? For sure, this exists in both first and second quadrants, but you are only interested in the half x = -√(y-y2). Since √(y-y2) ≥ 0, what quadrant does x=-√(y-y2) lie?
I agree on the bounds for θ, but the values for r suggest a circle of radius 1/2 centered on the origin. Consider the curve x2+y2=y. Convert this to polar coordinates and you will obtain r. This value of r will depend on θ.IsaacStats said:In the second quadrant = > (pi/2)<=theta<=pi
0<=r<=0.5
CAF123 said:I agree on the bounds for θ, but the values for r suggest a circle of radius 1/2 centered on the origin. Consider the curve x2+y2=y. Convert this to polar coordinates and you will obtain r. This value of r will depend on θ.
These bounds say that as θ goes from π/2 to π, r goes from 0 to 1. r is measured from the origin and so r is precisely 1 at only one value of θ, namely θ = π/2. (See this from a sketch). So the bounds for r have to depend on θ, otherwise you are describing a portion of the unit circle centered at the origin.IsaacStats said:Thank you for correction. If (pi/2)<=theta<=pi, does it meant that sin(pi)<=r<=sin(pi/2); 0<=r<=1.
that's weird
CAF123 said:These bounds say that as θ goes from π/2 to π, r goes from 0 to 1. r is measured from the origin and so r is precisely 1 at only one value of θ, namely θ = π/2. (See this from a sketch). So the bounds for r have to depend on θ, otherwise you are describing a portion of the unit circle centered at the origin.
Why did you tag on the part '0<=theta<=n/2, 0<=r<=1'? Remember x ≤ 0 so there should be nothing that describes anything in the first quadrant.IsaacStats said:That's great!
Does this mean that we can write the integral as a sum of two integrals (first quadrant circle and second quadrant circle)
0<=theta<=n/2, 0<=r<=1 + n/2<=theta<n, 0<=r<=sin(theta)
?
CAF123 said:Why did you tag on the part '0<=theta<=n/2, 0<=r<=1'? Remember x ≤ 0 so there should be nothing that describes anything in the first quadrant.
The bounds 'n/2<=theta<n, 0<=r<=sin(theta)' describe the required region. Did you draw a sketch? Convince yourself that this is the correct range for r - try out a few values of θ and see that you get reasonable values for r (r should monotonically decrease to 0 as θ sweeps from π/2 to π)
IsaacStats said:RIGHT! Function is undefined for X>0.
Therefore, n/2<=theta<=n, and 0<=r<=sin(theta)
CAF123 said:Yes, everything makes sense?
The curve in Cartesians was ##x^2 + y^2 = y##. Converting this to polar coordinates gives ##r^2 = r\sin\theta## from which we see that ##r = \sin \theta##. ##\theta = \pi/2## is the maximum value for ##r##, while ##\theta = \pi## is the minimum. In between, ##r## takes values given by the sine of the angle ##\theta##.IsaacStats said:Out of curiosity, why does not r go between 0 and csc(theta)?
CAF123 said:The curve in Cartesians was ##x^2 + y^2 = y##. Converting this to polar coordinates gives ##r^2 = r\sin\theta## from which we see that ##r = \sin \theta##. ##\theta = \pi/2## is the maximum value for ##r##, while ##\theta = \pi## is the minimum. In between, ##r## takes values given by the sine of the angle ##\theta##.
A double integral in Cartesian to Polar coordinates is a mathematical concept used to find the area under a surface in a two-dimensional space. It calculates the area by breaking it into smaller, rectangular pieces and adding them together. This method is useful for finding the area of shapes that cannot be easily calculated using traditional methods.
A single integral calculates the area under a curve in a one-dimensional space, while a double integral calculates the area under a surface in a two-dimensional space. In Cartesian to Polar coordinates, the double integral is calculated using two integrals, one for the radius and one for the angle.
The process for converting a double integral from Cartesian to Polar coordinates involves substituting the variables in the integrand with their polar coordinate equivalents. The limits of integration also need to be adjusted accordingly. The double integral can then be solved using the polar coordinate equivalents of the integration formulas.
Using a double integral in Cartesian to Polar coordinates allows for the calculation of complex areas that cannot be easily solved using traditional methods. It also allows for the integration of functions that are easier to express in polar coordinates, such as circles and spirals.
Double integrals in Cartesian to Polar coordinates are commonly used in physics, engineering, and other scientific fields to calculate the area and volume of complex shapes. They are also used in calculating moments of inertia and center of mass for objects with rotational symmetry.