Finding the area under the curve

[voidflux]

Hello everyone! i'm very confused on why i'm not getting this answer right, even the TI-83 says my answer is correct.
The directions say,
Consider the polar curve r = 2 sin(3x) for [0,PI]. Find the region inside the curve, using the calculator, i found the area as 1.3333, which is 4/3. I also found 4/3 by taking the definate integral from 0 to PI.

A = definate integral from 0 to PI (2*sin(3*x));
A = 2/3* definate integral from 0 to PI sin(u) du
2/3[-cos(3*x)] from 0 to PI.

i used u substitution u = 3x, du = 3dx, 1/3du = dx;

The only answers to choose from are:
2PI, PI, 3PI, 4PI, 5PI

and 4/3 isn't any of them!
any idea's?

[Something Else]

Originally posted by voidflux

The directions say,
Consider the polar curve r = 2 sin(3x) for [0,PI]. Find the region inside the curve

Is that what it says exactly?
or is there a presupposition of previous conditions established before you got to this part.

Anyways, no matter what, I can't help.

Too many years,
too many beers,
lost my interest in calculus.
-Lou

[voidflux]

Nope, thats exactly what the question says, for part A, then it says to find the slope at PI/4

[ZaidGS]

Originally posted by voidflux
Hello everyone! i'm very confused on why i'm not getting this answer right, even the TI-83 says my answer is correct.

if your calculator says you are correct u must be correct,
assuming u used it right.

yet i got one idea (although i dont really get ur question well)
a trick might lie in areas UNDER curve,
ie. if we got 3units ABOVE curve,
and 2units UNDER curve,
the answer is 1unit,
although the question might want 5,

so check out the co-ordinates of cutting x-axis.

good luck

[voidflux]

Alright thanks for the help! i'll see what I can do!

[alkatran]

Wow, my mind just started chugging when I heard that...

For some reason, I decided the best thing to do would be to put it in as a loop, calculating each point and the next (taking the average area under of the two * distance between). Just lower the distance between (increasing iterations) to increase the precision, increase precision until you get it to as many decimals as you need.

[prog_tom]

Originally posted by voidflux
Hello everyone! i'm very confused on why i'm not getting this answer right, even the TI-83 says my answer is correct.
The directions say,
Consider the polar curve r = 2 sin(3x) for [0,PI]. Find the region inside the curve, using the calculator, i found the area as 1.3333, which is 4/3. I also found 4/3 by taking the definate integral from 0 to PI.

A = definate integral from 0 to PI (2*sin(3*x));
A = 2/3* definate integral from 0 to PI sin(u) du
2/3[-cos(3*x)] from 0 to PI.

i used u substitution u = 3x, du = 3dx, 1/3du = dx;

The only answers to choose from are:
2PI, PI, 3PI, 4PI, 5PI

and 4/3 isn't any of them!
any idea's?

R = -2/3cos(3x)

R = (2/3) - (-2/3) = 4/3 in [0,PI]

[voidflux]

Ahh i c! so my answer is correct and all the other answers on his paper are incorrect then?
Thanks for the reply!