Question

write the definite integral that represents the area of the surface formed by revolving the graph f(x)=x^(1/2) on the interval [0,4] about the y-axis (Do not evaluate the integral.)

Answer 1

can you help me with the steps to get to that answer? i don't understand how you got there

Answer 2

The surface area can be considered the sum of the circumferences of an infinite number of circles:
The circumference of one such circle is \(2\pi r\), where \(r\) is the radius of each circle.

The integral would then look something like this:
\[A=2\pi\int_a^b(\text{radius})~ds\]
Here, \(ds=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}~dx\). I'm not as familiar with the derivation for \(ds\), but this link should explain it: http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

The radius of each circle is the distance from the curve to the axis of revolution. In this case, the axis is the y-axis, or \(x=0\). The distance from any given point on the curve to this axis is thus \(x\).