Express the volume of the solid D, enclosed by the surfaces z = √(x² + y²) and sphere of diameter 4 in cylindrical coordinates.

Steps to Solve

1. Convert to Cylindrical Coordinates In cylindrical coordinates, we have the transformations:

• ( x = r \cos(\theta) )
• ( y = r \sin(\theta) )
• ( z = z )

The equations for the surfaces become:

• The cone: ( z = r )
• The sphere of diameter 4: The radius is 2, so the equation is ( x^2 + y^2 + z^2 = 4 ), or ( r^2 + z^2 = 4 ).

2. Determine the Bounds for Integration The region ( D ) is enclosed by the cone and the sphere. We solve for the intersection of these two surfaces: Set ( z = r ) into the sphere's equation: $$ r^2 + r^2 = 4 $$ This simplifies to: $$ 2r^2 = 4 \implies r^2 = 2 \implies r = \sqrt{2} $$

Thus the limits for ( r ) are from 0 to ( \sqrt{2} ) and for ( z ) from ( r ) (the cone) up to the sphere.

3. Setting Up the Triple Integral The volume ( V ) can be calculated using the triple integral: $$ V = \int_0^{2\pi} \int_0^{\sqrt{2}} \int_r^{\sqrt{4 - r^2}} r , dz , dr , d\theta $$

4. Evaluate the Inner Integral First, evaluate the integral with respect to ( z ): $$ \int_r^{\sqrt{4 - r^2}} r , dz = r[z]_{r}^{\sqrt{4 - r^2}} = r(\sqrt{4 - r^2} - r) $$

5. Compute the Volume Integral Now substituting back into the remaining integrals: $$ V = \int_0^{2\pi} \int_0^{\sqrt{2}} r(\sqrt{4 - r^2} - r) , dr , d\theta $$

This consists of: $$ V = \int_0^{2\pi} d\theta \int_0^{\sqrt{2}} (r\sqrt{4 - r^2} - r^2) , dr $$

6. Final Integration Steps First, evaluate the integral ( \int_0^{\sqrt{2}} r\sqrt{4 - r^2} , dr ) and ( \int_0^{\sqrt{2}} r^2 , dr ).

Using substitution or looking up integral tables can help for the calculation of the first part.

The final result is obtained after performing all calculations and keeping track of constants.