Draw a circle in the x-z plane, radius 5 centred on the origin. This represents the outer sphere.
Draw two vertical lines, x=±3 and intersect the circle at 4 points. This represents the inner cylinder.
The points of intersections can be obtained by Pythagoras theorem to line on the lines z=±4. These will be the limits of integration for the required volume.
Now draw a horizontal elemental strip between the cylinder and the sphere, of height dz. This is the cross section of a hollow disk, outer radius,
R = sqrt(25-z^2)
and inner radius,
r = 3
The area of the elemental disk of height dz is:
dV=π(R^2-r^2)dz
and the volume of solid sought is:
V = ∫dV
=∫π(sqrt(25-z^2)dz
to be integrated from -4 to +4.
The indefinite integral is
%pi*(16*z-z^3/3), and when evaluated between -4 and +4,
V = (256π)/3 = 268.1 (approx.)
find the volume OUTSIDE the cylinder x^2+y^2=9 and INSIDE the sphere x^2+y^2+z^2=25
1 answer
1 answer