giantsace55

find the area of the region inside r=10sin(theta) but outside r=3

i get 154.08 as an answer but that doesnt seem to be correct

NewAccount

Is r=3 a vertical line or a horizontal line?

NewAccount

I'm getting 11.48 as the answer, assuming that r=3 is a horizontal line.

Graph the two functions on your calculator: y=10sin(x) and y=3.

You'll notice that they are asking for the area between these two curves. I'll be taking vertical strips, so the limits of integration are decided by the x coordinates of the intersection of the two curves. The two x coordinates are: .30469265 and 2.8369.

I now integrate (10sin(x) - 3) from .30469265 to 2.8369 and get 11.48.

Sound good?

giantsace55

hmm these are polar equations, isnt the formula for area of a polar equation integral (.5r^2)?

giantsace55

i found the answer, its 66.193

integral of 0.5 ((10 sin(x))^2-3^2) from x=.304 to 2.8369 dx = 66.2193

NewAccount

I'm a moron lol. My bad.