The base of a solid is the circle x^2+y^2=9. Cross sections of the solid perpendicular to the x-axis are semi-circles. What is the volume, in cubics units, of the solid?
a) 9 π/4
b) 18π
c) 9π
d) 72π
To find the volume of the solid, we need to integrate the area of each cross section perpendicular to the x-axis.
Since the cross sections are semi-circles, the radius of each semi-circle varies with the x-coordinate. From the equation of the circle x^2 + y^2 = 9, we can see that the radius is always 3 units, because the circle has a radius of 3 units.
Let's integrate the area of each semi-circle from x = -3 to x = 3.
The area of each semi-circle is given by A(x) = (1/2)πr^2, where r is the radius. In this case, r is 3 units.
Since the base of the solid is symmetric about the y-axis, we only need to evaluate the integral from x = 0 to x = 3, and then double the result.
So, the volume V of the solid is given by:
V = 2 * ∫[0 to 3] (1/2)π(3^2) dx
V = 2 * ∫[0 to 3] (9/2)π dx
V = 2 * (9/2)π * ∫[0 to 3] dx
V = 9π * [x] from 0 to 3
V = 9π * (3 - 0)
V = 27π cubic units
Therefore, the correct answer is:
c) 9π
Steve posted this solution in 2016 : )
each square of thickness dx has side 2y, so its area is 4y^2.
Adding up all the thin squares, and using symmetry,
v = 2∫[0,3] 4(9-x^2) dx
So just check the search engine for your question : ) Often it has been asked before.