What Shape is Formed by Revolving y = sqrt(a^2 - x^2) Around the X-axis?

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?Find the surface area when Y = (a^2 - x^2)^(1/2) is revolved around the x-axis?

?Geometrically, what have you found?

Please help. Thank you.
 
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Could you show us what you have tried? It might give us an idea on where you need help.

Also, there is a template for homework posts. For future reference, please use it. It should appear by default whenever you start a new topic in the homework section.
 
Char. Limit said:
Could you show us what you have tried? It might give us an idea on where you need help.

Also, there is a template for homework posts. For future reference, please use it. It should appear by default whenever you start a new topic in the homework section.

I just need help on how to set up the problem. (And i'll definitely use the template next time)
 
Well, firstly, how well do you know arc length? Because it will be needed here...

And tell me, how would you find the surface area of a regular solid, like say a cylinder, that has no top or bottom to consider?

Your equation in a more elegant format:

y=\sqrt{a^2-x^2}
 
calc II said:
?Find the surface area when Y = (a^2 - x^2)^(1/2) is revolved around the x-axis?

?Geometrically, what have you found?

Please help. Thank you.

Your equation y=\sqrt{a^2-x^2} may be more recognizable in the form y^2+x^2=a^2. What shape does the graph of this equation have? Can you deduce how y=\sqrt{a^2-x^2} relates to y^2+x^2=a^2? This should help with the second question (and also give you an idea of what to expect for the answer to the first question). As Char.Limit pointed out, a good look at the arc length and surface area formulas might be in order.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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