Question

Find the centroid of the region bounded by the given curves.
y = 6 sin 4x, y = 6 cos 4x, x = 0, x = pi/16

Answer 1

What have you tried? Did you set up the Three Integrals?

Answer 2

yeah i did. I worked it out and entered my answer but it was wrong.

Answer 3

Okay, what were your integrals?

Answer 4

\[\int\limits_{0}^{\pi/16} 6 \cos 4x - 6 \sin 4x dx \]

Answer 5

Okay, there's the base. I get numerically 0.621 or so.

How about the other two?

Answer 6

\[\int\limits_{0}^{\pi/16} 1/2* (6\cos 4x)^2 - (6\sin 4x)^2\]

Answer 7

\[\int\limits_{0}^{\pi/16} x* 6 \cos 4x - 6 \sin 4x\]

Answer 8

Hmmm... That's not quite it.

The first - Maybe you just didn't put the giant parentheses around the two squared monstrosities?

The second - I hope you mean \(x\cdot\left(6\cos(4x) - 6\sin(4x)\right)\).

Parentheses make all the difference.

Answer 9

yeah thats what I meant

Answer 10

How about the first one - the one with the squares? Both with 1/2?

Answer 11

\[Its \int\limits_{0}^{\pi/16} 1/A* 1/2* (6 \cos(4x)-6 \sin(4x))^2 dx \]

Answer 12

That's no good. Should be \(½6\cos^{2}(4x) - ½6\sin^{2}(4x)\)

Answer 13

alright thanks

Answer 14

so for my x bar- I have \[((3/32) \sqrt{2}\pi-4)/(3/2)(\sqrt{2}-1)\]

Answer 15

And my y bar is \[(3/8) /(3/2) (\sqrt{2}-1)\]

Answer 16

Well, that's ugly enough.

xbar is no good. Why is it negative?
ybar also no good.

Something going wrong in there.

Answer 17

when I get my Mx and My. what do i put it over?

Answer 18

The denominator is just the area of the region.

Should be: \(\dfrac{3}{32}\left(\pi\sqrt{2} - 4\right)\)

Looks like your close. Better algebra may be needed.

Answer 19

For the other, you should get simply 9/4.

Answer 20

The region area is just \(\dfrac{3}{2}\left(\sqrt{2} - 1\right)\)

Answer 21

Final Answer: \(\left(\dfrac{\pi\sqrt{2}-4}{16\cdot(\sqrt{2} - 1)},\dfrac{3}{2\cdot(\sqrt{2} - 1)}\right)\)