Question

The base of a solid is the circle x2 + y2 = 9. Cross sections of the solid perpendicular to the x-axis are semi-circles. What is the volume, in cubic units, of the solid?

Answer 1

Option b)
`18\pi` is correct

∴ the volume of the solid is
`A(x)=18\pi` cubic units

Explanation:

Given that the base of a solid is the circle
`x^2 + y^2 = 9` and Cross sections of the solid perpendicular to the x-axis are semi-circles.

To find the the volume of the solid in cubic units:

We know that the cross sections are semicircles with the diameter in the given circle
`x^2 + y^2 = 9`

That is we have to find the formula for the area of any semicircle perpendicular to x-axis, and integrate it from -3 to 3.

Now the area of a semicircle is

`A=(\pi r^2)/(2)` cubic units

Let r = y and
`y^2=9-x^2`

Then area of the semicircle crossing the x-axis at x is given by

`A(x)=(1)/(2)\pi y^2` cubic units

`=(1)/(2)\pi(9-x^2)`

Now we can find the definite integral of A(x) from x = -3 to x = 3.

Since A(x) is an EVEN function then the definite integral of A(x) from x = -3 to x = 3 is the same as twice the integral of A(x) from x = 0 to x = 3.

We have that

`A(x)=2(\int_0^3 (1)/(2)\pi(9-x^2))dx`

`=2((\pi)/(2)[9x-(x^3)/(3)]_0^3)`

`=\pi[9(3)-(3^3)/(3)-9(0)-(-(0^3)/(3))]`

`=\pi[27-(27)/(3)]`

`=\pi[27-9]`

`=\pi[18]`

`=18\pi`

∴ option b)
`18\pi` is correct

∴ the volume of the solid is
`A(x)=18\pi` cubic units