Question

find the area of the surface formed by revolving about the polar axis. r = 3cos(Theta) 0 11 years ago

Answer 1

hello

Answer 2

hey man thanks for coming to help me some this problem :)

Answer 3

*solve

Answer 4

i am still here btw

Answer 5

Set \(x=r\cos\theta=3\cos^2\theta\) and \(y=r\sin\theta=3\sin\theta\cos\theta\). Then
\[\begin{cases}
\dfrac{dx}{d\theta}=-3\sin2\theta\\\\
\dfrac{dy}{d\theta}=3\cos2\theta
\end{cases}\]
The arc length of a curve \(C\) in polar form is given by
\[\int_CdS=\int_C\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\,d\theta=\int_C\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta\]
The surface area is then given by
\[\int_C2\pi r\cos\theta\,dS=2\pi\int_0^{\pi/2}3\cos^2\theta\sqrt{9\sin^22\theta+9\cos^22\theta}\,d\theta\]

Answer 6

okay and i can factor out the 9's and the sin and cos turn into a 1 right?

Answer 7

Yes, the integral reduces quite a bit.
\[18\pi\int_0^{\pi/2}\cos^2\theta\,d\theta\]

Answer 8

okay and what happens when it revolves about the polar axis do i use the same equation

Answer 9

this other problem is asking to find the area of the surface formed by revolving about the polar axis so i am confused about what it meant by polar axis

Answer 10

The polar axis is the positive x-axis. That's already accounted for in the formula above with \(r\sin\theta\).

Speaking of which, the integral actually should be
\[\int_C2\pi r\color{red}{\sin}\theta\,dS=18\pi\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta\]
Sorry about that

Answer 11

okay so we still use that same formula of the surface area?

Answer 12

why do you have sin2(theta) and cos2(theta)?

Answer 13

Yes. If you're wondering about the derivation, it's all a matter of converting from rectangular to polar. Given a curve \(C\) defined over an interval \([a,b]\) by a curve \(\begin{cases}x=x(t)\\y=y(t)\end{cases}\), the area of the surface generated by revolving the curve about the x- or y-axis, respectively, is given by
\[A=2\pi\int_C y(x)\,dS\quad\text{or}\quad A=2\pi\int_Cx(y)\,dS\]
where \(dS\) is the arc length over the interval, given by the expression
\[dS=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\]
In converting to polar, we replace \(x=r\cos\theta\) and \(y=r\sin\theta\) and replace the derivatives accordingly with either \(\dfrac{dr}{d\theta}\) or \(\dfrac{dx}{d\theta}\) and \(\dfrac{dy}{d\theta}\).

I'm using trig identities. Differentiating \(x=r\cos\theta\) with respect to \(\theta\) gives
\[\frac{dx}{d\theta}=\frac{dr}{d\theta}\cos\theta-r\sin\theta\]
where \(r=3\cos\theta\), and so \(\dfrac{dr}{d\theta}=-3\sin\theta\). Recall that \(\sin2x=2\sin x\cos x\), and so
\[\frac{dx}{d\theta}=-3\sin\theta\cos\theta-3\cos\theta\sin\theta=-6\sin\theta\cos\theta=-3\sin2\theta\]

Answer 14

okay i see and when it says about the axis theta = pi/2

Answer 15

it says surface area formed by revolving r = e^6(theta) about the axis theta=pi/2 over the interval 0<theta<pi/2 do i go work it out the same way?

Answer 16

That would be the y-axis. In that case, you only exchange \(x\) for \(y\), which is actually the first integral setup I had:
\[18\pi\int_0^{\pi/2}\cos^2\theta\,d\theta\]

Answer 17

So this axis, \(\theta=\dfrac{\pi}{2}\), does it apply to a different problem?

Answer 18

yes sir

Answer 19

so just switch the x and y do i also change the sin with the cos?

Answer 20

That's what I mean: \(x\) corresponds to \(\cos\theta\) in the integral setup, and \(y\) to \(\sin\theta\).

Answer 21

For this next problem, you'd have
\[2\pi\int_0^{\pi/2}r\cos\theta\sqrt{\cdots}\,d\theta\]
where \(r=e^{6\cos\theta}\).

Answer 22

okay