Question

If the volume of the region bounded by the paraboloid $z = x^2 + y^2$ and the plane $z = 2y$ is given by \[ \int_{0}^{\alpha} \int_{\beta(y)}^{2y} \int_{-\sqrt{z - y^2}}^{\sqrt{z - y^2}} dx \, dz \, dy \] then

• A. $\alpha = 2$ and $\beta(y) = y, \, y \in [0,2]$
• B. $\alpha = 1$ and $\beta(y) = y^2, \, y \in [0,1]$
• C. $\alpha = 2$ and $\beta(y) = y^2, \, y \in [0,2]$
• D. $\alpha = 1$ and $\beta(y) = y, \, y \in [0,1]$

Hint:

For volume bounded by surfaces, identify intersection curves to determine correct $y$-limits before setting up the triple integral.

Answer 1

The Correct Option is C

Step 1: Find intersection curve

At the boundary: $$x^2 + y^2 = 2y$$ $$x^2 + y^2 - 2y = 0$$ $$x^2 + (y-1)^2 = 1$$

This is a circle of radius 1 centered at $(0, 1)$ in the $xy$-plane.

Step 2: Determine the region

The paraboloid opens upward from the origin. The plane $z = 2y$ is valid for $y \geq 0$.

For the region to exist (paraboloid below plane): $$x^2 + y^2 \leq 2y$$ $$x^2 + (y-1)^2 \leq 1$$

Step 3: Set up limits of integration

For a given $y$, the intersection circle gives: $$x^2 = 2y - y^2 = y(2-y)$$

This is real when $0 \leq y \leq 2$, and: $$-\sqrt{2y - y^2} \leq x \leq \sqrt{2y - y^2}$$

For $z$: At fixed $(x, y)$, $z$ ranges from the paraboloid to the plane: $$x^2 + y^2 \leq z \leq 2y$$

For $y$: The circle extends from $y = 0$ to $y = 2$ (diameter along $y$-axis from $(0,0)$ to $(0,2)$).

Step 4: Identify the integral

The volume integral is: $$V = \int_0^{\alpha} \int_{\beta(y)}^{2y} \int_{-\sqrt{z-y^2}}^{\sqrt{z-y^2}} dx,dz,dy$$

Wait, the given integral has limits: $$\int_0^{\alpha} \int_{\beta(y)}^{2y} \int_{-\sqrt{z-y^2}}^{\sqrt{z-y^2}} dx,dz,dy$$

Actually, looking at the given form more carefully: for fixed $y$, $z$ ranges from $\beta(y)$ to $2y$.

The lower bound for $z$ is the paraboloid: $z = x^2 + y^2$.

At the minimum (when $x = 0$): $z = y^2$

Therefore: $\beta(y) = y^2$

And: $\alpha = 2$ (upper limit for $y$)

Answer: (C) $\alpha = 2$ and $\beta(y) = y^2$, $y \in [0, 2]$