I need the area inside r=18sin(theta) but outside r=2
Im having trouble finding the interval on which to integrate.
Hello, casesam!
A sketch might help . . .Find the area inside , but outsideCode:18| * * * *:::::|:::::* *:::::::|:::::::* *::::::::|::::::::* :::::::::|::::::::: *:::::::::|:::::::::* *:::::::::|:::::::::* *:::::::::|:::::::::* :::::::ooo::::::: *:::o | o:::* *:o | o:* * | * -----o----*-*-*----o----- | 2 o | o o | o ooo |
The graphs intersect where: .
Hence: .
Due to the symmetry, we can integrate from 0.11134 to . . . and double.
Thanks for the graph! so now for the integration
(2)(1/2)(integral from 0.11134 to (pi/2)) (18sin(theta))^2-(2)^2
=(1)(integral) 324(sin(theta))^2-4
=(4)(integral) 81(sin(theta))^2-1
=(4)(integral)81((1/2)(1-cos(2theta)))-1
=(4)(integral) (81/2)-(81/2)cos(2theta)-1
=4[(79/2)theta-(81/4)sin(2theta)] integrated from 0.11134 to (pi/2)
Is this right? and what final answer do you get with many decimal places please.
Thanks in advance