Base of a solid

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?  

9x/4

18 pi

9π pi

72 pi

Answer 1

Gail,

 

Since the cross sections are semicircles with the diameter in the circle x² + y² = 9, what you need to do is find a formula for the area of any semicircle perpendicular to the x-axis, and integrate it from -3 to 3.

 

If you draw a chord across the x-axis from the bottom of this circle to the top, and say it crosses the x-axis at the number x, what is the length of the chord? The y-coordinate (height) from the x-axis to the circle will be the y-value for the value x obtained from the circle's equation. Since the chord is symmetric to the x-axis, its length will be 2y.

 

Now, what is the area of this semicircle? The area of a semicircle is πr²/2. Use r = y as I explained above. Then are area of the semicircle crossing the x-axis at x is

 

A(x) = 1/2 π y² = 1/2 π (9 - x²)    (notice that y² = 9 - x² is just a rearrangement of the equation of the circle).

 

to finish this, you can find the definite integral of A(x) from x = -3 to x = 3. If you use symmetry (A(x) is an EVEN function), this is the same as twice the integral of A(x) from x = 0 to x = 3.

 

This is not a hard integral to do, so I will just let you finish it yourself.