Find the area of the region described:

[hank]

Region is enclosed by the rose r = 4cos 3@.

I'm attempting to apply the formula:
S r^2 / 2 d@, with my limits 0 and 2 pi

So, here we go..

A = 1/2 S 16cos^2(3@) d@ , from 0 to 2pi
A = 4S(1 - cos6@) d@ //using trig identity to sub in for cos^2(3@)
A = 4[@ - sin(6@)/6], from 0 to 2pi
A= 8pi

However, according to the book it should be 4pi. All of my problems are like this where I end up with exactly twice the number I should have. This leads me to believe I'm missing a 1/2 from somewhere.

Can someone show me what I'm missing?

 

[skeeter]

the graph of \(\displaystyle \L r = 4\cos{(3\theta)}\) completes itself from 0 to pi ... in going from 0 to 2pi, you're tracing over the rose twice.

if you have a graphing calculator, graph the polar equation from 0 to 2pi and use the little "bug" tracer to the left of r₁ ... you'll see it travel the rose outline twice.

any rose with an odd multiple of theta will do this.

 

[hank]

If I were finding the area for a circle, would the same principle apply?

Because when I do it for a circle I end up getting 2x the answer I should be getting as well.

For instance, if r = 2asin@
A = 1/2 S 4a^2sin^2@ d@
= 2a^2 Ssin^2 * d@
= 2a^2 S 1/2 (1 - cos2@) d@
= a^2 S (1 - cos 2@) d@
= a^ 2 [@ - sin (2@) / 2] limits 0 to 2pi
= a^2 [2pi - 0 - (0 - 0)]
= 2pi * a^2

According to the book it is pi * a^2.

Do we only do it for 0 to pi for a circle as well?

 

[skeeter]

yes ... the circles \(\displaystyle r = a\cos{\theta}\) and \(\displaystyle r = a\sin{\theta}\) complete their respective graphs from 0 to pi ... in both cases, the coefficient of theta is 1, an odd number.

I recommend you get some polar graph paper and practice sketching some curves.

 

[hank]

Ok, thank you very much.

I will definately do that.