find area inside r=8sinΘ but outside r=4?
I've set r_1 = r_2
so 8sinΘ=4
this gives Θ=π/6
This gives me my upper limit of π/6
so then A=integral from 0 to π/6 of (8sinΘ)^2-4^2)dΘ
64sinΘ^2-16
I'm not really sure where to take it from here....any help would be great.
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You are on the right track, but there is a problem with your limits of integration. A good sketch would be very helpful here.
Yes, one of the intersection points of the curves occurs where Θ = π/6, but you made that the upper limit and placed the lower limit at Θ = 0. That is incorrect. There are two intersections. The limits are π/6 and 5π/6.
There is a problem with the integrand also. You correctly used the difference of the squares of the radii, but you must also divide that by 2. Here is the correct integrand:
[(8sinΘ)² - 4²]/2 = 32sin²Θ - 8
Another benefit of a good sketch is that you can see that the region is bounded by two circles, so a geometric solution is not too difficult, and it makes a good check.