Question

show that the area under r=a*sin（3theta) is 1/12 * a*pi, where theta is between 0 and 1/3 pi, inclusively

Answer 1

sorry, it should be 1/12*a^2*pi

Answer 2

@amistre64 help！！！!

Answer 3

\[\int^{\beta}_{\alpha}\frac {[r(t)]^2}2~dt\]

Answer 4

just remember the formula for the area of a circle:\[\frac12~2pi~r^2\]2pi represent the radians traveled, so replace 2pi with dt

Answer 5

do you recall the integration of sin^2(x) ?

Answer 6

nope， not at all

Answer 7

well, last time we tangoed ... you said you hadnt learnded by parts yet. Have you learnded it yet?

Answer 8

ye, I learn that but I don't know to use it on the sin^2 x function

Answer 9

\[s^2(3t)=s(3t)~s(3t)\]

u = s(3t) v = -c(3t)
du = 3 c(3t) dv = s(3t)

int s^2(3t) = -s(3t) c(3t) - int -3 c(3t) c(3t) dt

Answer 10

define your parts and run with it ... youll see that by the second run thru you are going in circles ..... but theres a way out of it

Answer 11

\[\int s^2(3t)~dt=-s(3t)~c(3t)/3-\int-3c(3t)~c(3t)/3~dt\]
\[\int s^2(3t)~dt=-s(3t)~c(3t)/3+\int c(3t)~c(3t)~dt\]
run thru by parts with the cosines now

u = c(3t) v = s(3t)/3
du = -3 s(3t) dv = c(3t)

\[\int s^2(3t)~dt=-s(3t)~c(3t)/3+\left(s(3t)~c(3t)/3-\int -3s(3t)~s(3t)/3~dt\right)\]
\[\int s^2(3t)~dt=-s(3t)~c(3t)/3+s(3t)~c(3t)/3+\int s^2(3t)~dt\]

pretty sure ive got an error in that someplace .... coding and mathing tend to mess things up for me :)

we should have a multiple of the original integral on the end so that we can subtract it from both sides

Answer 12

might have to work up some trig identity stuff ....

sin^2(3t) = 1-cos^2(3t)

cos(2x) = cos^2(x) - sin^2(x)
= cos^2(x) - (1 - cos^2(x))
= 2cos^2(x) - 1

(cos(2x) + 1)/2 = cos^2(x) , let x = 3t

-cos^2(3t) = -(cos(6t)+1)/2

\[\int sin^2(3t)dt=\int 1-\frac12cos(6t)~dt=t-\frac12~\frac16sin(6t)\]

Answer 13

so, this amounts to\[\frac12a^2\left( t-\frac1{12}sin(6t)-[t-\frac1{12}sin(6t)]\right)^{pi/3}_{0}\]

\[\frac12a^2\left( \frac{pi}3-\frac1{12}sin(6*\frac{pi}3)-[0-\frac1{12}sin(6.0)]\right)\]
\[\frac12a^2\left( \frac{pi}3-\frac1{12}(0)\right)\]

Answer 14

i missed the +1 when i coded up the bottom of the other post :/

Answer 15

\[\int sin^2(3t)dt=\int 1-\frac12(cos(6t)+1)~dt=t-\frac12~\frac16sin(6t)-\frac12t\]
\[\frac12t-\frac1{12}sin(6t)\]