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Area under polar curve

Untitled.png

Part (b) I know how to find the area under the curves, but I'm not sure what limits should be used and what should be subtracted from what. The points of intersection are (1,π6) (1, \frac{\pi}{6}) and (1,5π6)(1,\frac{5\pi}{6})
(edited 7 years ago)
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B_9710
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Original post by NotNotBatman
Untitled.png

Part (b) I know how to find the area under the curves, but I'm not sure what limits should be used and what should be subtracted from what. The points of intersection are (1,π6) (1, \frac{\pi}{6}) and (1,5π6)(1,\frac{5\pi}{6})


Polar.png
Find the area of the red region by using the coordinates that you have found, then add it to the area of the black region which is a sector.
Original post by B_9710
Polar.png
Find the area of the red region by using the coordinates that you have found, then add it to the area of the black region which is a sector.


The area of the red region is ∫π6π2r2dθ \int_{\frac{\pi}{6}}^\frac{\pi}{2}r^2 d\theta?
Original post by NotNotBatman
The area of the red region is ∫π6π2r2dθ \int_{\frac{\pi}{6}}^\frac{\pi}{2}r^2 d\theta?


Yes. Should there be a 1/2 in front of that?? Polar knowledge is dim with me nowadays.
Original post by RDKGames
Yes. Should there be a 1/2 in front of that?? Polar knowledge is dim with me nowadays.


I did the red area individually and multiplied by 2, I hope.
(edited 7 years ago)
Original post by NotNotBatman
I did the 2 sectors individually and multiplied by 2, I hope.


Oh then yeah if you're doing both red's at once then it will be that.
Reply 6
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B_9710
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Original post by NotNotBatman
I did the 2 sectors individually and multiplied by 2, I hope.


Yeah there is two of them so it's fine, no need for the 1/2, it's better this way.
Original post by B_9710
Yeah there is two of them so it's fine, no need for the 1/2, it's better this way.


Original post by RDKGames
Oh then yeah if you're doing both red's at once then it will be that.


Thank you, the diagram made it much easier.

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