Can we get a negative value for areas of surfaces of revolution?

joseph9496
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the question is..

Find the areas of the surfaces generated by revolvin the curves about the indicated axes.

x = (1/3)y^(3/2) - y^(1/2), 1≦y≦3; revolved about y-axis

so i use the general formula "S = Integral 2π (radius)(dS)"
and the radiu in this case is x which is (1/3)y^(3/2) - y^(1/2) but yea..
i just keep getting negative value when i subsitute everything in the equation...

thanks.
 
Last edited:
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joseph9496 said:
the question is..

Find the areas of the surfaces generated by revolvin the curves about the indicated axes.

x = (1/3)y^(3/2) - y^(1/2), 1 <= y <= 3; revolved about y-axis

so i use then general formula "S = 2π INTEGRAL (radius)(dS)"
and the radiu in this case is x which is (1/3)y^(3/2) - y^(1/2) but yea..
i just keep getting negative value when i subsitute everything in the equation...

thanks.

Yes, the surface area is given by 2\pi \int x dS. Here dS is the differential of arc length, \sqrt{1+ (dx/dy)^2} dy
Since x= (1/3)y^(3/2)- y^(1/2), dx/dy= (1/2)y^{1/2}- (1/2)y^{-1/2} and (dx/dy)^2= (1/4)(y- 2+ y^{-1}). Then (dx/dy)^2+ 1= (1/4)(y+ 2+ y^{-1})= (1/4)(y^{1/2}+ y^{-1/2})^2 so dS= \sqrt{(1/4)(y+ 2+ y^{-1})}= (1/2)(y^{1/2}+ y^{-1/2})dy.

The surface area is
\pi\int_1^3 (y^{1/2}+ y^{-1/2}) dy
That should be easy to integrate and you definitely won't get a negative value!

To answer you original question, no "area" is never negative.
 
HallsofIvy said:
The surface area is
\pi\int_1^3 (y^{1/2}+ y^{-1/2}) dy
That should be easy to integrate and you definitely won't get a negative value!

Yes i got to that part too but what happen to the radius (x)?
cuz after i multiply the radius with the dS and subsitute value 3 and 1 into y-value i got
-16/9..
 
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You are right- I forgot to multiply by x= (1/3)y^{3/2}- y^{1/2}. That would make the integral
\pi \int_1^3 ((1/3)y^2- (2/3)y- 1)dy
 
yea..and after i do anti-derivative i got

π [ (1/9)y^3 - (2/6)y^2 -y ]

which gives me a -16/9...

i just dun understand how come my answer is negative..

btw thanks for helping!
 

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