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: Identify the surfaces.
We have two surfaces: \[ z = x^2 + y^2 \text{and} z = 2y. \] They intersect when $x^2 + y^2 = 2y$, i.e., $x^2 + (y - 1)^2 = 1$. This is a circle centered at $(0,1)$ with radius $1$. Thus $y \in [0,2]$.

Step 2: Determine limits of $y$ and $z$.
For each $y$, $z$ ranges from $z = x^2 + y^2$ (lower surface) to $z = 2y$ (upper surface). Also, from the integrand, $x$ ranges from $-\sqrt{z - y^2}$ to $\sqrt{z - y^2}$.

Step 3: Find correct bounds.
Since $z = 2y$ intersects $z = x^2 + y^2$ at $y=0$ and $y=1$, we take $y \in [0,1]$. Hence $\alpha = 1$ and $\beta(y) = y^2$.

Step 4: Conclusion.
\[ \boxed{\alpha = 1, \, \beta(y) = y^2, \, y \in [0,1]}. \]