The goal is to determine the correct limits $a$, $b$, and $c$ for the given triple integral representing the volume of the described solid. Let's analyze the problem step by step.

1. Understanding the problem:

• The top surface is given by $z = 6 - x^2 - y^2$, which is a paraboloid opening downwards.
• The lateral boundary is given by $x^2 = 2 - y^2$, which is a circle in the $xy$-plane.
• The bottom boundary is at $z = 0$, representing the $xy$-plane.

We will convert the given problem into cylindrical coordinates: $x = r\cos\theta, \quad y = r\sin\theta, \quad z = z$ $x^2 + y^2 = r^2$

The equation $z = 6 - x^2 - y^2$ becomes $z = 6 - r^2$, and $x^2 = 2 - y^2$ becomes $r^2 = 2$.

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2. Converting the limits:

• Angular limit ($a$):
  The paraboloid and circle are symmetric about the origin, so $\theta$ ranges from $0$ to $2\pi$:
  $a = 2\pi$

• Radial limit ($b$):
  The circle in the $xy$-plane is $x^2 + y^2 = r^2 = 2$, so $r$ ranges from $0$ to $\sqrt{2}$:
  $b = \sqrt{2}$

• Vertical (z) limit ($c$):
  The paraboloid extends from $z = 0$ to $z = 6 - r^2$:
  $c = 6 - r^2$

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3. Correct answer:

The limits are: $a = 2\pi, \quad b = \sqrt{2}, \quad c = 6 - r^2$

Thus, the correct option is: $\boxed{a = 2\pi, b = \sqrt{2}, c = 6 - r^2.}$

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Do you have any questions or need further clarification? 😊

Related Questions:

1. How do we find the volume of the solid using these limits?
2. Why is the angular limit $a$ exactly $2\pi$?
3. What is the significance of the equation $x^2 + y^2 = r^2$ in cylindrical coordinates?
4. Can this problem be solved using spherical coordinates instead of cylindrical coordinates?
5. How do we graph $z = 6 - r^2$ and $r^2 = 2$ to visualize the solid?

Tip:

Always sketch the region of integration to better understand the bounds and their relationships in cylindrical or spherical coordinates!