Revolution of a curve along a axis

kazthehack
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Homework Statement


given z =y3 revolved around the y-axis what is the equation of the surface and then graph.


Homework Equations





The Attempt at a Solution


I have no idea how to approach this, i tried searching around the net but nothing came out. pls help.
 
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start with the equation of a circle of radius z = y^3, with centre on the y-axis & in a plane perpindicular to the y axis
 
^ what i need is a step by step solution and i learn to solve other problems if i see how it is done.
 
sorry, you have to try - I'm happy to walk you through it if you attempt, but I'm not just going to do it all for you
 
kazthehack said:
given z =y3 revolved around the y-axis what is the equation of the surface and then graph.
…
I have no idea how to approach this, i tried searching around the net but nothing came out. pls help.

Hi kazthehack! :wink:

Start with a single point z =y03 (for a fixed value y0) revolved around the y-axis …

what is the equation of that? :smile:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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