ARandomOWl
New Member
I have a set of parametric equations given by:
[LATEX]x=t-2sin t[/LATEX]
[LATEX]y=1-2cos t[/LATEX]
[LATEX]0\leqslant t \leqslant 2 \pi[/LATEX]
When y=0: [LATEX]t=\frac{\pi}{3}, \frac{5 \pi}{3}[/LATEX]
The question: Show that the area enclosed by the curve and the x-axis is given by the integral:
[LATEX]\int_{\frac{\pi}{3}}^{\frac{5 \pi}{3}} (1-2cost)^2dt[/LATEX]
Can someone explain why the area is not given by:
[LATEX]\int_{\frac{\pi}{3}}^{\frac{5 \pi}{3}} (1-2cost)dt[/LATEX]
I noticed that:
[LATEX](1-2cost)^2 = y^2 = y \frac{dx}{dt}[/LATEX]
But I don't know of it's significance.
Thanks.
[LATEX]x=t-2sin t[/LATEX]
[LATEX]y=1-2cos t[/LATEX]
[LATEX]0\leqslant t \leqslant 2 \pi[/LATEX]
When y=0: [LATEX]t=\frac{\pi}{3}, \frac{5 \pi}{3}[/LATEX]
The question: Show that the area enclosed by the curve and the x-axis is given by the integral:
[LATEX]\int_{\frac{\pi}{3}}^{\frac{5 \pi}{3}} (1-2cost)^2dt[/LATEX]
Can someone explain why the area is not given by:
[LATEX]\int_{\frac{\pi}{3}}^{\frac{5 \pi}{3}} (1-2cost)dt[/LATEX]
I noticed that:
[LATEX](1-2cost)^2 = y^2 = y \frac{dx}{dt}[/LATEX]
But I don't know of it's significance.
Thanks.
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