Question

Find the area of the surface obtained by rotating the curve x= 2cos(t)-cos(2t) and y=2sin(t)- sin(2t) about the x-axis

the shape I get is completely wrong to do a problem like this

Answer 1

\[S=\int_\alpha^\beta 2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\]
Are the bounds on t really not given?

Answer 2

If not, then you need to deduce them from the equation for y. To do this, we need to find the x-intercepts of the parametric equations:
\[\begin{align}
y=0 &\Rightarrow 2 \sin(t)-\sin(2t)=0\\
&\Rightarrow 2 \sin(t)-2\sin(t)\cos(t)=0\\
&\Rightarrow (1-\cos(t))\sin(t)=0\\
&\Rightarrow 1-\cos(t)=0 \text{ OR } \sin(t)=0
\end{align}\]

Answer 3

One obvious solution is \(t=0\) and the nearest solution to this is \(t=\pi\). Thus, take \(\alpha=0\) and \(\beta=\pi\).

Answer 4

Wait I need to deduce them? I thought it was from the y formula

Answer 5

I did use the y-formula.

Answer 6

I meant x, woah sorry about that I'm tired. I just dont get the shape, when I sketch it, I get a continuous function kinda similar looking to sin(t) and cos(t)

Answer 7

so then I just have to find the maximum from 0? I mean the function doesn't limit itself at 0 for y.

Answer 8

You can use WolframAlpha to draw the shape for you, and it looks like it's a cardioid with x-intercepts x=0 and x=-3. This agrees with my conclusion that \(t=0\) and \(t=\pi\) are the bounds for the integral.

Answer 9

Sorry this answer is kinda late, however how does that help you find the limits if x is 0 and -3? do you just plug in the intercept in the y euqation? Plus I used graph sketcher and it showed something else ( dunno if it will show): http://www.graphsketch.com/?eqn1_color=5&eqn1_eqn=2cos(x)-cos(2x)&eqn2_color=2&eqn2_eqn=2sin(x)-sin(2x)&eqn3_color=3&eqn3_eqn=&eqn4_color=4&eqn4_eqn=&eqn5_color=6&eqn5_eqn=&eqn6_color=6&eqn6_eqn=&x_min=-17&x_max=17&y_min=-10.5&y_max=10.5&x_tick=1&y_tick=1&x_label_freq=5&y_label_freq=5&do_grid=0&do_grid=1&bold_labeled_lines=0&bold_labeled_lines=1&line_width=4&image_w=850&image_h=525

Answer 10

You should enter it under "Parametric", not "Functions".
For the first question, you can plug x=1 (it's supposed to be 1, not 0, my bad) and x=-3 in the equation for x.
For example, if x=1, we have \(2 \cos(t)-\cos(2t)=1\). If you simplify this, you get \[2 \cos(t)-(2 \cos^2(t)-1)=1 \Rightarrow 2 \cos^2(t)-2\cos(t)=0\] and from this, you can see that t=0 will satisfy this equation.

Answer 11

Yeah I solved for it, however the integral seems a bit complicated to solve for , the derivative within the square root is giving me a hard time, guess I'll have to ask the professor.

Answer 12

You can simplify the expression under the root to 1-cos(t).

Answer 13

waiit I start up with (2sint + 2sin2t)^2+ (2cost-2cos2t)^2 after the expansion I end up with about 8 different terms that I have to use trig identities for?

Answer 14

You can simplify most of it since there are 2 pairs of \(\sin^2\) and \(\cos^2\) terms. Expand patiently and you will see that you can simplify a lot of it.

Answer 15

BTW, it's supposed to be (-2sin(t)+2sin(2t))^2.

Answer 16

ahhh alright thank you man! thanks for your patience

Answer 17

No problem, dude. :)