find the volume OUTSIDE the cylinder x^2+y^2=9 and INSIDE the sphere x^2+y^2+z^2=25

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.)

To find the volume outside the cylinder and inside the sphere, we need to calculate the volume of the remaining solid region.

Step 1: Determine the limits of integration for the solid region.
Let's consider the xy-plane as the base of the solid. To find the limits of integration for x and y, we need to determine the intersection points between the cylinder and the sphere.

First, equate the equations of the cylinder and sphere to find the intersection points:
x^2 + y^2 = 9 (equation of the cylinder)
x^2 + y^2 + z^2 = 25 (equation of the sphere)

Subtracting the equation of the cylinder from the sphere gives:
z^2 = 25 - 9 = 16

Taking the square root, we have:
z = ±4

So, the intersection points are (0, 0, ±4).

Step 2: Determine the limits of integration for x and y.
The cylinder has a radius of √9 = 3. Thus, the limits of integration for x and y are -3 to 3.

Step 3: Set up the triple integral.
The volume of the solid region can be calculated through a triple integral.

V = ∫∫∫ dV

Where dV represents an infinitesimal volume element.

Step 4: Set up the limits of integration.
The limits of integration for x and y are -3 to 3, as determined in step 2.
The limit of integration for z is from the sphere's surface (z = 4) to the cylinder (z = 0).

V = ∫[-3, 3]∫[-3, 3]∫[0, 4] dz dy dx

Step 5: Integrate.
Integrating the triple integral, we have:

V = ∫[-3, 3]∫[-3, 3]∫[0, 4] dz dy dx

Integrating with respect to z gives:

V = ∫[-3, 3]∫[-3, 3] [4z] [z = 0 to z = 4] dy dx

V = ∫[-3, 3]∫[-3, 3] (16) dy dx

Integrating with respect to y gives:

V = ∫[-3, 3] [16y] [y = -3 to y = 3] dx

V = ∫[-3, 3] (96) dx

Integrating with respect to x gives:

V = [96x] [x = -3 to x = 3]

V = 96(3) - 96(-3)

V = 288 + 288

V = 576 cubic units.

Therefore, the volume outside the cylinder and inside the sphere is 576 cubic units.

To find the volume outside the cylinder and inside the sphere, we need to calculate the volume of the sphere and subtract the volume of the cylinder from it.

The equation of the cylinder is x^2 + y^2 = 9, which represents a cylinder with radius 3 and axis along the z-axis.

The equation of the sphere is x^2 + y^2 + z^2 = 25, which represents a sphere with radius 5 centered at the origin.

To find the volume of the sphere, we can use the formula for the volume of a sphere: V_sphere = (4/3) * π * r^3, where r is the radius of the sphere.

Substituting the radius, we get V_sphere = (4/3) * π * 5^3 = (4/3) * π * 125 = 523.6 cubic units (rounded to one decimal place).

To find the volume of the cylinder, we can use the formula for the volume of a cylinder: V_cylinder = π * r^2 * h, where r is the radius of the base and h is the height of the cylinder.

In this case, the height of the cylinder extends from -√(25 - 9) = -√16 = -4 to √(25 - 9) = √16 = 4, as the cylinder is bounded by the surface of the sphere.

Substituting the radius and height, we get V_cylinder = π * 3^2 * 8 = 72π cubic units.

Finally, to find the volume outside the cylinder and inside the sphere, we subtract the volume of the cylinder from the volume of the sphere:

V_outside_cylinder_inside_sphere = V_sphere - V_cylinder = 523.6 - 72π cubic units.

Note: This result is an exact expression in terms of π and cannot be simplified further without more specific values.