Surface Area of a curve rotated about an axis

[jmshiraef]

I need to find the area of the surface obtained by rotating the curve about the x-axis.

The equation for the curve is x = ((y^2-ln[y])/(2*sqrt[2])

or x = (one over two times the square root of two) multiplied by (y squared minus the natural log of y)

So far I have found dx/dy to be (2y - y^(-1))/(2*sqrt[2])
and 1 + (dx/dy)^2 = 1/8(4y^2 + y^(-2) - 1/2)

I am not sure what to do next and have been stuck for a very long time.

Any help you can provide would be greatly appreciated.

Thanks,

 

[tkhunny]

The whole thing? Surely there was some range of values suggested.

Where's the square root of the whole mess?

 

[galactus]

Your integral is:

\(\displaystyle \L\\\frac{\pi}{4}\int_{a}^{b}\frac{(y^{2}-ln(y))(2y^{2}+1)}{y}dy\), \(\displaystyle a,b > 0\)

Now, it is your mission to get to this point and then integrate. But first, you need limits of integration.

 

[jmshiraef]

The limits are 1 and 5 for y

 

[Deleted member 4993]

Break-up the expression that galactus gave you - and integrate. It is tedious - but not hard.