Question

Calc 2 Polar Equations Question:
What is the area from 0 to pi/2 of r = sin(2θ)

Answer 1

\[\Large Area=\int\limits_{0}^{\frac{\pi}{2}}\sqrt{r^2 + (\frac{dr}{d \theta})^2}d \theta \] \[\Large r = sin(2 \theta )\] \[\Large (\frac{dr}{d \theta})^2 = 2cos(2 \theta)\]
So \[\Large Area=\int\limits_{0}^{\frac{\pi}{2}}\sqrt{sin^2(2 \theta ) + (2cos(2 \theta))^2}d \theta \]
And at this point I can't figure out how to manipulate the equation into a workable integral... :/

Answer 2

That's not the area integral. It should be
\[A = \int_a^b \frac{1}{2} r(\theta)^2 d\theta \]

Answer 3

What you have there is the arc length integral.

Answer 4

doh. >.< That would probably make things a lot easier. Thanks.