# Arc length, area

1. Mar 24, 2010

### lockedup

1. The problem statement, all variables and given/known data
For the curve $$y=\sqrt{x}$$ , between x = 0and x = 2, find (a) the area under the curve, (b) the arc length, (c) the volume of the solid generated when the area is revolved about the x axis, (d) the curved area of this solid.

2. Relevant equations
$$ds = \sqrt {1+(y')^{2}}dx$$

3. The attempt at a solutionI did a and c pretty easily. My problem is arc length. I wrote out the integral $$\int^{2}_{0} \sqrt{1+\frac{1}{4x}}dx$$ and realized I don't have a clue how to calculate it.

Also, for the "curved" area. Does my professor mean surface area? How do I do that?

2. Mar 24, 2010

### Dick

To integrate that try a substitution like u^2=1+1/(4x). That will make it a rational function and then you can use partial fractions and stuff. If you don't 'know' how to solve a given integral, that's usually your first move. And the 'curved surface area' is area swept out by the curve. Look up 'surface of revolution'. There's a very similar formula to the one you have for arc length.