To solve this problem, let's analyze the limits of integration required for finding the volume of the solid region $E$ between the two paraboloids $z = x^2 + y^2$ and $z = 18 - x^2 - y^2$.

Step 1: Understanding the Equations of the Paraboloids

1. The first paraboloid is given by $z = x^2 + y^2$.
2. The second paraboloid is given by $z = 18 - x^2 - y^2$.

Step 2: Converting to Polar Coordinates

In polar coordinates:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $x^2 + y^2 = r^2$

Thus, the equations of the paraboloids become:

1. $z = r^2$
2. $z = 18 - r^2$

Step 3: Finding the Intersection of the Paraboloids

To determine the bounds for $r$, we set the two equations equal to each other to find where they intersect: $r^2 = 18 - r^2$ Solving for $r^2$, we get: $2r^2 = 18 \Rightarrow r^2 = 9 \Rightarrow r = 3$ Thus, $r$ ranges from $0$ to $3$.

Step 4: Determining the Limits of Integration

From the given integral form $V(E) = \int_0^{2\pi} \int_0^a \int_b^c r \, dz \, dr \, d\theta$, we identify the limits as follows:

1. $\theta$: Since the region is rotationally symmetric around the $z$-axis, $\theta$ ranges from $0$ to $2\pi$.

   • So, $a = 2\pi$.

2. $r$: The radial distance $r$ ranges from $0$ to $3$ (the intersection point of the paraboloids).

   • So, $c = 3$.

3. $z$: For a fixed $r$, $z$ ranges from the lower paraboloid $z = r^2$ to the upper paraboloid $z = 18 - r^2$.

   • So, $b = r^2$ and $c = 18 - r^2$.

Final Answer:

• $a = 2\pi$
• $b = r^2$
• $c = 18 - r^2$

Would you like further details or have any questions?

Follow-Up Questions

1. How do polar coordinates simplify volume calculations in symmetric solids?
2. Why is the volume integral set up with three nested integrals?
3. How would this setup change if the region were not symmetric around the z-axis?
4. Can the volume of other shapes be calculated using similar integral setups?
5. How would one solve the integral explicitly for the volume of the region?

Tip:

When dealing with solids of revolution, switching to polar or cylindrical coordinates often simplifies the integral setup by leveraging symmetry.