anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
blockcolder
  • blockcolder
\[S=\int_\alpha^\beta 2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\] Are the bounds on t really not given?
blockcolder
  • blockcolder
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: \[\begin{align} y=0 &\Rightarrow 2 \sin(t)-\sin(2t)=0\\ &\Rightarrow 2 \sin(t)-2\sin(t)\cos(t)=0\\ &\Rightarrow (1-\cos(t))\sin(t)=0\\ &\Rightarrow 1-\cos(t)=0 \text{ OR } \sin(t)=0 \end{align}\]
blockcolder
  • blockcolder
One obvious solution is \(t=0\) and the nearest solution to this is \(t=\pi\). Thus, take \(\alpha=0\) and \(\beta=\pi\).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Wait I need to deduce them? I thought it was from the y formula
blockcolder
  • blockcolder
I did use the y-formula.
anonymous
  • anonymous
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)
anonymous
  • anonymous
so then I just have to find the maximum from 0? I mean the function doesn't limit itself at 0 for y.
blockcolder
  • blockcolder
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.
anonymous
  • anonymous
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
blockcolder
  • blockcolder
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.
anonymous
  • anonymous
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.
blockcolder
  • blockcolder
You can simplify the expression under the root to 1-cos(t).
anonymous
  • anonymous
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?
blockcolder
  • blockcolder
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.
blockcolder
  • blockcolder
BTW, it's supposed to be (-2sin(t)+2sin(2t))^2.
anonymous
  • anonymous
ahhh alright thank you man! thanks for your patience
blockcolder
  • blockcolder
No problem, dude. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.