Question

K, can someone check my work I'm trying to find the surface area of a polar coordinate system

Answer 1

Problem

r=sin2theta

Answer 2

\[1/2\int\limits_{0}^{\pi/2} \sin^22\theta \]

Answer 3

\[1/2\int\limits_{0}^{\pi/2} (1-\cos4\theta)d \theta\]

Answer 4

\[1/2(\theta - 1/2 \sin4\theta)\]

Answer 5

\[1/2(\pi/2-1/2\sin4(\pi/2) - 0 \]

Answer 6

= pi/4 ? does everything look good

Answer 7

hmmm... i keep getting pi/8 for the area of that 1 leaf in the first quadrant....

Answer 8

i don't beleive you u-sub'd correctly. you have:
\[\huge \int\limits \cos(4\theta) d \theta\]
but that should be \[\huge \int\limits {1 \over 4}\sin(4 \theta)\]
right?

Answer 9

thats right my mistake thanks!

Answer 10

my bad, i typo'd that. Should just be 1/4*sin(4x), no integral sign. My bad.