area of polar curve of r= 2+sin(4x)

need help finding the limits of integration

Question

area of polar curve of r= 2+sin(4x)

need help finding the limits of integration

anonymous · Answer

the maximum value of sin would be "1" when this happens, r = "3"
so, "r" goes from 0 to 3
sin(4x) has a period of $\pi/2$

anonymous · Answer

if i use the bounds of 0 and 2pi/4. i always get 1/4 the right answer. but shouldnt i get the complete right answer since i integrated from 0 to 360 degrees. upper limit is 2pi/4 because sin(4x) so the 4's cancel

anonymous · Answer

$$A=\int_{r=0}^3\int_{x=0}^{\pi/2} r\,dr\,dx$$

anonymous · Answer

ok

anonymous · Answer

no. if you over-trace the angle, your area becomes a numerical multiple.

anonymous · Answer

how did i over trace though?

anonymous · Answer

infact, you should integrate only half of it and multiply it by "2"
so, the correct way would be
$$A=2\int_{r=0}^3\int_{x=0}^{\pi/4}r\,dr\,dx$$

anonymous · Answer

you completed a period and going over the same path again.. so, you are computing the area of the region that you've already covered,