Consider a solid with a base defined by the function f(x) = x^2 on the interval [0, 3], and square cross sections perpendicular to the x-axis. What is the integral setup to find the volume of this solid?

A function f(x) = sqrt(x) is revolved around the x-axis from x = 0 to x = 4. What is the integral setup to find the volume of the resulting solid of revolution?

Let $R$ be the region bounded by $y=2-x^2,x=0$, and $y=0$ in the first quadrant. Use the shell method to find the volume of the solid generated when $R$ is revolved about the $y$-axis.

What is the surface area of the solid formed by revolving $f\left(x\right)=\cos x$ on $[0,\frac{\pi }{3}]$ about the $x$-axis?

Calculate the arc length of the line $y=-x+4$ on the interval $[2,7]$ using calculus, and verify with the geometric distance.