Surface Area of Rotated Curve y=1+5x2 | x=0 to x=8

danielatha4
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Homework Statement


Find the area of the surface obtained by rotating the curve y=1+5x2 from x=0 to x=8 about the y-axis



Homework Equations


As=\int2\pig(x)sqrt(1+[g'(x)]2)


The Attempt at a Solution


I changed the limits to suit the y direction, the lower limit becomes 1 and the upper 321. I solved x as a function of y in order to rotate around the y-axis and when I plug everything into the formula is becoems a really messy integral I can't solve. I'm sure there's an easier way.
 
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Hi danielatha4! :smile:

Why not just use ∫ 2πx √(1 + g'(x)2) dx ?
 

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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