surface area by rotating the curve.. work shown please help

[johnq2k7]

Surface area obtained by rotating the curve x= 2-y^2 around the y-axis

work shown:

y^2= 2-x
y=sqrt(2-x)

dy/dx=-1/(2*sqrt(2-x))

sqrt(1+(dy/dx)^2)= sqrt{(7-4x)/(8-4x)}

therefore, integral of 2*Pi*x*[sqrt(7-4x)/8-4x)]dx from 0 to r should provide the surface area of the curve rotated around the y-axis

is this correct?.. I think i made a few mistakes please help

 

[galactus]

Surface area obtained by rotating the curve x= 2-y^2 around the y-axis

\(\displaystyle {\pi}\int_{0}^{2}x\sqrt{\frac{4x-9}{x-2}}dx\)

 

[fasteddie65]

Why not use the disk method?

dV = ?x^2 dy = ? (2 - y^2)^2 dy

Use symmetry. 1/2 of the volume would be from y = 0 to y = whatever the upper limit is.
(You don't tell us what that is. I am guessing it is y = ?2, if you want to rotate the area in the first quadrant.

 

[galactus]

Be careful, this is a SURFACE of revolution, not a volume of revolution.

Yes, since this is surface area and we are revolving about the y-axis, we could use:

\(\displaystyle 2{\pi}\int_{0}^{\sqrt{2}}(2-y^{2})\sqrt{1+4y^{2}}dy\)

You should get the same solution as in my first post. Same thing.