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

1. Jan 29, 2008

### joseph9496

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: Jan 29, 2008
2. Jan 29, 2008

### HallsofIvy

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.

3. Jan 29, 2008

### joseph9496

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..

Last edited: Jan 29, 2008
4. Jan 30, 2008

### HallsofIvy

You are right- I forgot to multiply by $x= (1/3)y^{3/2}- y^{1/2}$. That would make the integral
$$\pi \int_1^3 ((1/3)y^2- (2/3)y- 1)dy$$

5. Jan 30, 2008

### joseph9496

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!