Yes, that is valid because the shaded region is symmetric wrt the y axis.PsychonautQQ said:Should I integrate from the intersection point to pi/2?
I would suggest subbing in arcsin(2/9). The range of this function is [-pi/2,pi/2] so subbing this in will give you the required theta.9sin(theta) = 2
theta = arcsin(2/9)
theta = .224093
Yes.If I do that that is half of the area and then I can multiply that by two to get the full area. Does that look better to you?
I plug these in for theta and then get a required theta? I'm confused by what you mean here. Do you mean that the limits of integration are [-∏/2,∏/2]?CAF123 said:I would suggest subbing in arcsin(2/9). The range of this function is [-pi/2,pi/2] so subbing this in will give you the required theta.
Yes.
No, just sub in arcsin(2/9) instead of subbing in 0.224... to avoid rounding errors. The principal solution of arcsin(2/9) is the one that appears in [-pi/2,pi/2] which is the theta that the two curves intersect at in the first quadrant.PsychonautQQ said:I plug these in for theta and then get a required theta? I'm confused by what you mean here. Do you mean that the limits of integration are [-∏/2,∏/2]?
The formula for finding the area between two polar curves is given by the integral of 1/2 times the difference between the outer and inner curves, squared, with respect to theta. This can be represented as: A = 1/2 ∫(r_outer^2 - r_inner^2) dθ.
The limits of integration for finding the area between polar curves can be determined by finding the points of intersection between the two curves. These points will serve as the limits for the integral with respect to theta.
No, the technique for finding the area between polar curves is different from that used for Cartesian curves. This is because polar coordinates use a different coordinate system and require the use of polar equations and integration with respect to theta.
Yes, it is possible to find the area between three or more polar curves by using multiple integrals. The first integral would involve finding the area between the outermost curve and the innermost curve. Then, this result would be used in a second integral to find the area between the next outermost curve and the previous innermost curve, and so on.
Yes, polar curves can be used to find the area of irregular shapes. By using multiple polar equations, we can divide the irregular shape into smaller, more manageable areas and then add them together to find the total area of the shape.