<p>The region enclosed by the x-axis, the line x=3 and the curve y = sqrt(x) is rotated about the x-axis. What is the volume of the solid generated. Anyone have an explanation? Thanks!</p>
<p>Using disk method, the radius of the disks when you rotate it around the x-axis would be R(x)=sqrt(x). Use the formula Volume = pi<em>fnInt(R(x)^2, x, 0, 3) to approximate the area. That would become pi</em>4.5 or 14.14</p>
<p>sorry if youre confused, it's kinda hard to do calculus on the computer</p>
<p>mr. stevezilla,</p>
<p>fnInt(R(x)^2, x, 0, 3) why is it r(x)^2? because pi(r^2)(dx) = V?</p>
<p>Also, would you mind if I asked you one more question?</p>
<p>Well here is the other problem if anyone else wants to jump in:</p>
<p>Let R be the region bounded by the graph of f(x) = 4x^2 - x^3 and the x-axis. Find the volume of the solid generated when R is revolved about the x-axis.</p>
<p>I'm not sure which bounds to choose for the integral, but I guess you would do... fnInt(f(x)^2,x,0,?) Thanks all for the help.</p>
<p>-Mike</p>
<p>Solve for the zeros. 0 and 4 I think. So do integral from zero to four of (pi(the function)^2). That should be right.</p>