Swayum
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A curve is described by

x = a\cos^3t,       y = asin^3t,       0 \leq  t \leq \pi /2

The curve is rotated through 2pi radians about the x-axis. Find the exact value of the area of the curved surface generated.

I've done the integration and all that but I'm confused about the limits I should use. A picture of the curve is attached. When it cuts the x axis, y = 0, so

asin^3t = 0
sin^3t = 0
sint = 0
t = 0 (pi is not a solution)

And the curve cuts the y axis when x = 0, so

cost = 0
t = pi/2

Then shouldn't the limits of the integral be \int^0_{\pi/2} f(t) \mathrm{d}t? Why is it the other way round? I'm sure it's something obvious that I'm missing, it is 6 AM after all....
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ahezhara
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Hmm. I've done this Q successfully. Hangon.
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ahezhara
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My GUESS is...

The formula for the cartesian curved surface area is ... well, the one with 1+\frac{dy}{dx} in it (sorry my Latex skills are bad!)

The limits for that will be in increasing order e.g. 0 and 1.

When you derive the formula for the same thing in parametric form, it's like a substitution for t instead of x, and with a substitution comes a change in limits, so x value goes to t value and this may warrant a swap-over to decreasing order.

Well... I might be being stupid, it's 6.30AM or so after all!
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gyrase
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No, thats not how you work out the limits.

The limits are given in the question. Upper bound pi/2, lower bound 0.
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ahezhara
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Yes, but it hasn't stated which is the upper and which is the lower bound.
Anyway, this is a stab in the dark, it does seem odd but if you're going in order of increasing x values then the t will reverse its values in this case.
I'm guessing that the formula for curved surface area is derived in cartesian form first (though I may be wrong) and so when specifying the limits it goes in order of increasing x values, not t values.

Edit: Thinking about it more, and I'm pretty sure that's it.
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gyrase
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(Original post by ahezhara)
Yes, but it hasn't stated which is the upper and which is the lower bound.
Anyway, this is a stab in the dark, it does seem odd but if you're going in order of increasing x values then the t will reverse its values in this case.
I'm guessing that the formula for curved surface area is derived in cartesian form first (though I may be wrong) and so when specifying the limits it goes in order of increasing x values, not t values.

Edit: Thinking about it more, and I'm pretty sure that's it.
It doesnt have to state it;

Definition: If a function y = f(x) has a continuous first derivative throughout the interval a < x < b, then the area of the surface generated by revolving the curve about the x-axis is; S = 2pi INT y(1+(dy/dx)^2)^(1/2).dx [from a to b]

source: http://curvebank.calstatela.edu/arearev/arearev.htm
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DoMakeSayThink
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 \int_{a}^b f(x) dx = -\int_{b}^a f(x) dx, and as area is scalar, it shouldn't make a big difference.
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Zii
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(Original post by Swayum)
A curve is described by

x = a\cos^3t,       y = asin^3t,       0 \leq  t \leq \pi /2

The curve is rotated through 2pi radians about the x-axis. Find the exact value of the area of the curved surface generated.

I've done the integration and all that but I'm confused about the limits I should use. A picture of the curve is attached. When it cuts the x axis, y = 0, so

asin^3t = 0
sin^3t = 0
sint = 0
t = 0 (pi is not a solution)

And the curve cuts the y axis when x = 0, so

cost = 0
t = pi/2

Then shouldn't the limits of the integral be \int^0_{\pi/2} f(t) \mathrm{d}t? Why is it the other way round? I'm sure it's something obvious that I'm missing, it is 6 AM after all....
It's given in parametric form, so use

2 \pi \displaystyle\int^{t=b}_{t=a} y\sqrt{\left(\frac{\mathrm{d}x}{  \mathrm{d}t} \right)^2 + \left(\frac{\mathrm{d}y}{\mathrm  {d}t} \right)^2} \, \mathrm{d}t

Everything must be in terms of t (including y, and the limits). The limits are in terms of t, not x, and so you must use the limits a = 0, b = \frac{\pi}{2}.

The formula I gave is in the formula book. If it's parametric, make sure you go with parametric.

The way I do these questions is I work out either what \sqrt{1+ \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2} is (if it's cartesian, not parametric) or I work out what\sqrt{(\frac{\mathrm{d}x}{\mathr  m{d}t})^2 + (\frac{\mathrm{d}y}{\mathrm{d}t}  )^2} is, if it is given in parameters.
In the question given, I found this to be 3a\mathrm{sin}t\mathrm{cos}t (after fumbling around, cancelling down, manipulation, etc), and so I then multiplied this by y = a\mathrm{sin}^3t and thus the integral becomes 6 \pi a^2\displaystyle\int^{\frac{\pi}  {2}}_0 \mathrm{sin}^4t\mathrm{cos}t \, \mathrm{d}t which is easy to solve.
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The Sherminator
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Okay for this question the limits are given in the question,  0.5 \pi and 0.
Its a standard result, you just need the formula, which is the integral of  2\pi y ds
It should work from there.

Btw, those values are the "t" values. If you used a sub like I did, then in the end I got limits of 0 and 1. So it should be clear where each one goes.
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