Find the volume:

Question

Find the volume of the solid whose base is the region between the inverted parabola y= 4-x^2 and the x-axis, and whose vertical cross sections perpendicular to the y axis are semi circles.

Answer 1

Let's start with the formula for the volume of a circular cylinder:  V = pi*r^2*h.  Cutting this in half in a direction parallel to the height produces a semicircular cylinder, where V = (pi/2)*r^2*h.  Since h is in the direction of the y-axis, we can find the differential volume by replacing h with dy, and we get

 

dV = (pi/2)*r^2*dy.

 

Now, we need to imagine the area we're integrating over, and how it relates to the radius.  The parabola intercepts the x-axis at x=-2 and x=2.  The y-axis is its line of symmetry.  And the radius of our differential volume can be drawn in the xy plane horizontally from the y-axis to either side of the parabola.  This means that the x-values of the parabola form the radii of the semicircles. 

 

So, we solve y = 4-x^2 for x, which gives us x = ±√(4-y).  Since the horizontal distance from the y-axis to either the positive or negative square root is the same, and both are equal to the radius (a positive quantity), let's choose

x = √(4-y). 

 

Noting that |x| and r are the same, we can re-write this as

 

r = √(4-y).

 

Substituting this into the formula for the volume differential, we get

 

dV = (pi/2)*[√(4-y)]^2*dy, or

dV = (pi/2)*(4-y)dy.

 

Since the integral is with respect to y, we find the lower and upper limits of integration by looking at the graph of the region in the xy-plane, and taking its lowest and greatest values for y.  We see that y ranges from its value at the x-axis, which is 0, to its value at the vertex of the parabola, which is 4.

 

Now, we integrate, from y=0 to y=4, the expression we have for dV. I'll put the limits of integration before the integral symbol so they don't get confused with the function we're integrating:

 

(from y=0 to 4)∫(pi/2)*(4-y)dy.