Volume of region bounded by two functions

Question

You may use a calculator and round to the nearest thousandth.

Let the region R be the area enclosed by the function f(x)=x^2, the horizontal line y=3, and the y-axis. If the region R is the base of a solid such that each cross-section perpendicular to the x-axis is a semi-circle with diameters extending through the region R, find the volume of the solid. You may use a calculator and round to the nearest thousandth.

Answer 1

First we need to find the points where the two functions intersect:

3-x² = 0

subtract 3 from both sides

x² = 3

square root both sides

x = √3

Then we have the two intersection points of the functions:

(0,0) and (√3,3)

Next we need the formula for the area of a semicircle:

A = πr²/2

r = 3-x² this is the radius of the circle at the point (√3,3)

Now substitute this r into the formula for the area of a semicircle:

A = π(3-x²)²/2

Now we can use calculus to integrate this area from x=0 to x=√3

= ∫π(3-x²)²/2 dx, from x=0 to x=√3

= π/2•∫(3-x²)² dx, from x=0 to x=√3

= π/2•∫(3-x²)(3-x²)dx, from x=0 to x=√3

= π/2•∫9-6x²+x⁴ dx, from x=0 to x=√3

= π/2[9x-6x³/3+x⁵/5], evaluated from x=0 to x=√3

= 13.059