Find the volume of the solid whose base is the region bounded by y=x^2 and the line y=0 and whose cross sections perpendicular to the base and parallel to the x-axis are semicircles.
Find the volume of the solid whose base is the region bounded between the curve y = √x and the x-axis from x = 0 to x = 4 and whose cross sections taken perpendicular to the x-axis are squares.
To find the volume of the solid, we'll use the method of cylindrical shells.
First, let's find the limits of integration for x. We can find the intersection points of the parabola y = x^2 and the line y = 0. Setting the equation equal to each other, we get x^2 = 0, which implies x = 0. Therefore, the region of integration will be bounded by x = 0 and some value x = a.
Next, let's find the height of each semicircular cross section. We know that the cross sections are parallel to the x-axis, so the height will be determined by the function y = x^2.
The formula for the volume of a cylindrical shell is V = ∫(2πrh)dx, where r is the radius of the shell, and h is the height of the shell. In this case, the radius of each semicircle will be half of the width of the region, which is x. So the radius r will be x/2, and the height h will be the function y = x^2.
Therefore, the volume of the solid will be V = ∫(2π(x/2)(x^2))dx, from x = 0 to x = a.
Simplifying, we get V = π∫(x^3)dx, from x = 0 to x = a.
Integrating, we get V = π[(1/4)x^4] from 0 to a.
Evaluating the integral, we get V = π[(1/4)(a^4) - (1/4)(0^4)], which simplifies to V = π(a^4)/4.
Therefore, the volume of the solid whose base is the region bounded by y = x^2 and y = 0, and whose cross sections are semicircles, is given by V = π(a^4)/4.