<p>find the area of the region inside r=10sin(theta) but outside r=3</p>
<p>i get 154.08 as an answer but that doesnt seem to be correct</p>
<p>Is r=3 a vertical line or a horizontal line?</p>
<p>I’m getting 11.48 as the answer, assuming that r=3 is a horizontal line. </p>
<p>Graph the two functions on your calculator: y=10sin(x) and y=3. </p>
<p>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.</p>
<p>I now integrate (10sin(x) - 3) from .30469265 to 2.8369 and get 11.48.</p>
<p>Sound good?</p>
<p>hmm these are polar equations, isnt the formula for area of a polar equation integral (.5r^2)?</p>
<p>i found the answer, its 66.193</p>
<p>integral of 0.5 ((10 sin(x))^2-3^2) from x=.304 to 2.8369 dx = 66.2193</p>
<p>I’m a moron lol. My bad.</p>