The elemental area of an isosceles triangle subtending an angle of δθ and radius(length) r is
(1/2)r²δθ.
The required area is therefore:
∫(1/2)r²δθ
for θ between the given limits (0 orπ?) to 1.5π.
You will need to substitute r by 2θ as in the given function.
Find the area of the region which is bounded by the polar curves
theta =pi and
r=2theta 0<theta<1.5pi inclusive
2 answers
As I noted in my original answer, there is an inconsistency in the specified limits of integration or "boundaries" of your enclosed area.
The spiral curve r = 2 theta starts at theta = 0 and ends at theta = 1.5 pi. The straight line theta = pi is the -x axis. It is not clear how the area from theta = pi to theta = 3/2 pi should be treated when doing the area integration.
You have been told how to do the integration. Try various integration limits (0 to pi; 0 to 3/2 pi; pi to 3/2 pi) and see which your grading software accepts.
The spiral curve r = 2 theta starts at theta = 0 and ends at theta = 1.5 pi. The straight line theta = pi is the -x axis. It is not clear how the area from theta = pi to theta = 3/2 pi should be treated when doing the area integration.
You have been told how to do the integration. Try various integration limits (0 to pi; 0 to 3/2 pi; pi to 3/2 pi) and see which your grading software accepts.
3 answers