Well, actually, Dick, area in polar coordinates is r dtheta! You didn't say quite what you meant to, did you?Dick said:Who said everything has to be super hard? But if you work that out, you'll get 0 for the area. Is that right? It might help to draw a picture. And, hey, area in polar coordinates isn't the integral of r dtheta, is it?
Would you look up the right formula for area?
HallsofIvy said:Well, actually, Dick, area in polar coordinates is r dtheta! You didn't say quite what you meant to, did you?
The equation for finding the area enclosed by r=3sin theta is A = 1/2 * r^2 * theta, where r is the radius and theta is the angle in radians.
The radius can be found by simply plugging in the given value for theta into the equation r=3sin theta. In this case, the radius would be 3.
The unit of measurement for the area enclosed by r=3sin theta would be square units, such as square inches or square meters, depending on the context of the problem.
No, the area enclosed by r=3sin theta cannot be negative since it is a measurement of area which cannot be negative. However, the value of r or theta can be negative, which would change the direction of the area but not its magnitude.
To find the area enclosed by r=3sin theta for a specific interval of theta, you would first find the angle in radians for the starting and ending points of the interval. Then, you would plug these values into the equation A = 1/2 * r^2 * theta and subtract the area at the starting point from the area at the ending point to find the area enclosed by the interval.