Integrating a Polar Equation

[tomcant]

I have been pulling my hair out for the last 30 minutes or so. In a practice maths paper I have, the first question is to find the the area enclosed by one loop of the curve, r = 2sin(3x) (where x is theta; I couldn't find the key for theta). The answer given in the answer booklet is pi / 24, but I'm sure this is wrong, unless I'm missing something. Would someone check this for me?

1/2 integral( 4sin^2(3x) ) dx
2 integral( 1/2 - 1/2cos(6x) ) dx
integral( 1 - cos(6x) ) dx

[ x - 1/6sin(6x) ] from 0 to pi/3

=> pi /3

which is not equal to pi/24 !! Is the answer booklet wrong?

[tomcant]

Anyone? This is really important, I have my final exam on monday!

[Jetru]

Is it sin^2(3x) or just sin(3x)?

[tomcant]

The curve to be integrated is r = 2sin(3x). So we integrate 1/2 r^2 dx, which is 1/2 (4sin^2(3x)) dx.

[Jetru]

That works only for polynomials. For trigonometric functions the integrals are different. Do some research.

Hint:Derivatives are inverses of integrals. To check if your integral is correct differentiate it and you should get the same function.

Edit: Okay something is funny here. The answer according to me is 4/3. I have no idea how pi comes into the answer. Or perhaps the the function is in polar form? r=2*sin 3x?

[tomcant]

What?! You must be confused! I know how to integrate cartesian equations, but this is not a cartesian equation. r = 2sin(2x) is a polar equation, hence the `r' (I use `x' because I don't know what the keyboard combo for theta is).

[edit]: Saw your edit. Yes, it is polar, which is why I get a multiple of pi.

[Darobat]

Say hello to theta
θ
And as an added bonus, I'll throw in pi!
π

Pi looks really dumb in this font.

[Jetru]

Can't find anything wrong. Probably a mistake in the book.

[tomcant]

Thanks! Thats what I wanted to hear.

[Jetru]

Wait, shouldn't we remove that extra area?

[tomcant]

That extra area isn't there. In the region 0 to pi/2, r=0 in two places; x=0 and x=pi/3. Where did you get pi/6 from? We should be integrating from 0 to pi/3.

[Jetru]

Well the graph is a polar graph. Here's a graph.


The area under the curve upto x axis is also counted. And the area can't be removed because we are taking a square over r. Everything get's added. At pi/6 ,the curve is at isn't outermost point. So the area under that upto x-axis is taken. Then we move backward and the area upto x-axis is taken again. From the above integral(that you calculated), we need to subtract twice the same integral from 0 to pi/6. But I do that and still don't get the correct answer.

[tomcant]

I see what you're saying, but I still disagree. To find the area of that loop, we must integrate between the two values of x for which r = 0. They are 0 and pi/3. I'm pretty certain that the area under the graph will not be counted (only the enclosed area will be counted).

Anyway, I had my final exam today and I think it went really well! I even had time to count up what I definately got right, which I think is about 90%. So I'm very happy right now! The only thing I know I got wrong was: "sinh(x) + 2cosh(x) = 8. Solve for x". I know this is easy, but I got all the way to finding the roots of a quadratic when instead of dividing by 10 (in the quadratic formula), I divided by 2! Nevermind...

Thanks for the help, Jetru.


[edit]: By the way, your calulator looks very similiar to mine, its not the "CFX-9850G PLUS" by any chance?

[Jetru]

Nope, it's the TI-86. Old one. Mathematica wasn't working correctly so I had to use my calc emulator.

[tomcant]

Did you pay for your copy of Mathematica? Or did you get it through your education? I really, really want a copy and I was wondering if I can get money off since I'm a student.