The document discusses three applications of integral calculus: 1) Finding the area between two curves by integrating the difference of the functions over a given interval. 2) Computing the volume of a solid obtained by rotating a curve around the x-axis using the disk method. 3) Calculating the volume of a solid generated by revolving an area bounded by two curves around the x-axis using the ring method.

1. Applications of Integral Calculus
The Area between Two Curves
Find the area between and from x = 0 to x = 1.
To get the height representative rectangle in the Figure, subtract the y-coordinate of its bottom from the y-
coordinate of its top. Its base is infinitesimal dx. So, because area equals height times base,
[( ) ]
Now, just add up the areas of all the rectangles from 0 to 1 by integrating:
∫ [( ) ]
[ ]
( ) ( )
Solids of Revolution: Circular Disk Method
Find the volume of the solid – between x = 2 and x = 3 – generated by rotating the curve y = ex about the
x-axis.

2. Each cross-section is a circle with a radius of ex. So, its area is given by the formula for the area
Plugging ex into r gives us: ( ) .
Adding up the volumes from 2 to 3 and integrating:
∫
[ ]
[ ]
Solids of Revolution: Circular Ring Method
Take the area bounded by and √ , and generate a solid by revolving that area about the x-
axis.

3. It should take a very little trial and error to see that and √ intersect at x = 0 and x = 1.
The area of the ring is equal to the area of the entire circle minus the area of the hole.
The area of the circle minus the hole is , where R is the outer radius (the big radius) and r is
the radius of the hole (the little radius). For this problem, the outer radius is √ and the radius of the hole
is , so that gives us (√ ) ( )
Multiply this area by the thickness, dx:
∫
∫
[ ]
[( ) ( )]
V = 0.94 cubic units
Reference: Calculus for Dummies by Mark Ryan © Wiley Pub., 2003
Elements of Calculus and Analytic Geometry by Francisco G. Reyes and Jenny L. Chua © 1992 UST
Publishing House Manila, Philippines.