To set up an integral in polar coordinates to find the volume of the ice cream cone-shaped solid bounded by the surfaces $z = \sqrt{x^2 + y^2}$ and $z = \sqrt{18 - x^2 - y^2}$, we can proceed as follows:

Step 1: Convert to Polar Coordinates

In polar coordinates, we have:

• $x = r \cos(\theta)$
• $y = r \sin(\theta)$
• $z = r$ (from $z = \sqrt{x^2 + y^2}$)

The second boundary, $z = \sqrt{18 - x^2 - y^2}$, becomes $z = \sqrt{18 - r^2}$.

Step 2: Volume Integral Setup

The volume $V$ can be found by integrating the function $z$ (height of the solid) over the region in the $r$-$\theta$ plane. The volume element $dV$ in cylindrical coordinates is $r \, dr \, d\theta \, dz$.

Since $z$ ranges from $r$ (the lower surface) to $\sqrt{18 - r^2}$ (the upper surface), we set up the triple integral as follows: $V = \int_0^{2\pi} \int_0^{\sqrt{18}} \int_r^{\sqrt{18 - r^2}} r \, dz \, dr \, d\theta$

Step 3: Evaluate the Integrals

1. Innermost Integral (in terms of $z$): $\int_r^{\sqrt{18 - r^2}} r \, dz = r \left[ z \right]_{z = r}^{z = \sqrt{18 - r^2}} = r \left( \sqrt{18 - r^2} - r \right)$

2. Remaining Integral (in terms of $r$ and $\theta$): Now the volume integral becomes: $V = \int_0^{2\pi} \int_0^{\sqrt{18}} r \left( \sqrt{18 - r^2} - r \right) \, dr \, d\theta$

Step 4: Set the Limits and Order of Integration

Given the instructions, we can match the required integral setup as follows:

• A (lower limit for $\theta$): $0$
• B (upper limit for $\theta$): $2\pi$
• C (lower limit for $r$): $0$
• D (upper limit for $r$): $\sqrt{18}$

The integrand is $r \left( \sqrt{18 - r^2} - r \right)$, and the order of integration is $dr \, d\theta$.

Final Answer

To fill in the blanks:

• Integrand: $r \left( \sqrt{18 - r^2} - r \right)$
• $dr$
• $d\theta$
• $A = 0$
• $B = 2\pi$
• $C = 0$
• $D = \sqrt{18}$

Would you like me to evaluate the integral to find the exact volume?

Related Questions to Explore

1. How would the integral setup change if the cone's orientation were different?
2. What if the cone had a different upper boundary, like $z = k - r^2$?
3. Can we compute the volume using spherical coordinates instead?
4. How does the setup vary if we switch the order of integration?
5. What geometric insights can we draw about volumes bounded by surfaces in polar coordinates?

Tip: When setting up integrals in polar coordinates, always visualize the radial limits in terms of the problem's symmetry and region of interest.