Find the volume of the solid whose base is enclosed by the circle...?

Question

Find the volume of the solid whose base is enclosed by the circle
x^2 + y^2 = 4 and whose cross sections taken perpendicular to the x-axis are squares.

V=?
Thanks! :)

Answer 1

Its not a cylinder. The cross-sections of a cylinder are rectangles. I will solve this generally for any radius. The base is then defined by:

x^2 + y^2 = r^2

Solve for the limits on y: y = sqrt(r^2 - x^2)
ymin = -sqrt(r^2 - x^2)
ymax = sqrt(r^2 - x^2)

So the y width is 2sqrt(r^2 - x^2)

Since the cross sections are all squares, the height equlas the width. Set up a differential volume measuring the width by the height by dx.

dV = (2 sqrt(r^2 - x^2))(2 sqrt(r^2 - x^2)) dx = 4(r^2 - x^2) dx

Now integrate: V = 4(xr^2 - (x^3)/3) 4x(r^2 - (x^2)/3

Apply the limits of -r to + r: V = 8r(2(r^2)/3) = (16r^3)/3

In your case r = 2 so V = 128/3

Answer 2

x^2+y^2=4 is the equation of a circle with radius 2 or diameter of 4
since cross section is square, the height is equal to diameter i.e. 4
Volume is pi*r^2*height
pi*4*4
16pi

Answer 3

that's a cylinder with a base radius of 2 and height of 4

V = (Area of Base) * height

Solve for area of circle : pi * 2^2
multiply by 4