It is? i don't see that. Since r > 0 for all theta, the curve cannot join up until it has gone all the way around 2π.CompuChip said:Hhint: the correct answer is related to the period of r(theta)).
CompuChip said:Exactly. So why the pi / 4?
The equation for finding the area enclosed by a polar curve is A = 1/2 ∫ab r(θ)^2 dθ, where r(θ) is the polar curve equation and a and b represent the limits of integration.
The limits of integration can be found by setting the polar curve equation equal to 0 and solving for θ. These values will represent the starting and ending points of the area enclosed by the curve.
No, the area enclosed by a polar curve is always positive. It represents the magnitude of the enclosed region and does not take into account direction.
To graph a polar curve, first identify the symmetry and important points of the curve. Then, plot points by substituting values of θ into the polar curve equation. Finally, connect the points to create the graph of the curve.
If the polar curve intersects itself, the area enclosed can be found by breaking the curve into separate regions and finding the area of each region individually. Then, add the areas together to find the total enclosed area.