How Do You Calculate the Area Bounded by \( r = 8\cos(10\Theta) \)?

Cici2006
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Homework Statement


Find the area of the region bounded by r=8cos10\Theta


Homework Equations





The Attempt at a Solution



I set r=0 to find \Theta, which i used for my bounds
\Theta=pi/20, 3pi/20
A= \int(1/2)64cos^2(10\Theta) d\Theta
 
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What you need to do is multiply your answer by 20 since you found the area of one of the 20 petals of the rose curve.
 
Are you having trouble finding the correct solution since yours is too small? or because you don't know how to evaluate the integral?

If you need help evaluating the integral, use the fact that

\cos^2{nx} = \frac{1+\cos{2nx}}{2}, n\in\mathbb{N}
 
Okay, let me state what i did in more detail:
A=(1/2)integral 64(cos^2(10theta)) d(theta)
=32 integral (1/2)(1+cos20theta) (theta)
=16[theta-(1/20)sin20theta]
did i do it correct so far?
then i just plug in my bounds which are pi/20 to 3pi/20 right?
now should i just multiply my answer by 20?
 
thanks for the help i solved it
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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