Problem : Apply the disk method to find the volume of the unit sphere obtained by revolving the region below the graph of f (x) = from x = - 1 to 1 about the x -axis.

Solution for Problem 1 >>

Using the disk method, we have

Π ()2 dx	=	Π (1 - x 2)dx
	=	Π -
	=

as expected.

Close

Problem : Apply the shell method to find the volume of the solid obtained by revolving the region below the graph of f (x) = x ² from x = 0 to 1 about the y -axis (this solid looks like a cylinder with a bowl carved out of the top).

Solution for Problem 2 >>

The formula from the shell method gives

2Π x(x 2)dx	=	2Π x 3 dx
	=	2Π |0 1
	=

Close

Problem : Compute the volume of the square pyramid with base in the x = 0 plane with sides of length 10 and vertex at the point (5, 0) , using the cross-sectional area method.

Solution for Problem 3 >>

The cross-section of the pyramid in the plane perpendicular to the x -axis with a particular x -coordinate in the interval [0, 5] is a square with sides of length s(x) = 10 - 2x . Such a square has area equal to A(x) = s(x)² = (10 - 2x)² = 4x ² - 40x + 100 , so the volume of the pyramid is given by

A(x)dx	=	(4x 2 - 40x + 100)dx
	=	-20x 2 + 100x
	=

Close