Volume Between Paraboloids z=x²+y² and z=4-x²-y²

The image depicts a solid Region $R$ in three-dimensional space, enclosed by two intersecting paraboloids: one opening upward from the origin ($z = x^2 + y^2$) and one opening downward from $z=4$ ($z = 4 - x^2 - y^2$). The intersection of these surfaces forms a circle of radius $\sqrt{2}$ at the height $z=2$, which projects onto the $xy$-plane as the disk $x^2 + y^2 \leq 2$.

Answer

The volume of the region $R$ is calculated by integrating the difference between the upper paraboloid and the lower paraboloid over the circular domain in the $xy$-plane. Using cylindrical coordinates, the volume is found to be $4\pi$.

Explanation

1. Setting up the Triple Integral The volume $V$ of a solid region $R$ is defined by the triple integral over the constant function 1. Based on the boundaries provided, we set the $z$-limits from the lower paraboloid to the upper paraboloid. $V = \iiint_{R} dV = \iint_{D} \left( \int_{x^2+y^2}^{4-x^2-y^2} dz \right) dA$ This expression calculates the vertical height at every point $(x,y)$ within the projection $D$.

2. Converting to Cylindrical Coordinates Given the circular symmetry about the $z$-axis, we use cylindrical coordinates where $x^2 + y^2 = r^2$. The Jacobian of the transformation is $r$, so $dV = r \, dz \, dr \, d\theta$. $V = \int_{0}^{2\pi} \int_{0}^{\sqrt{2}} \int_{r^2}^{4-r^2} r \, dz \, dr \, d\theta$ The radial limit is $\sqrt{2}$ because the intersection occurs where $r^2 = 4 - r^2 \implies 2r^2 = 4 \implies r = \sqrt{2}$.

3. Evaluating the Inner Integral We first integrate with respect to $z$, which represents the thickness of the solid at a specific radius $r$. $\int_{r^2}^{4-r^2} dz = (4 - r^2) - r^2 = 4 - 2r^2$ This formula represents the vertical distance between the top and bottom surfaces as a function of the radius.

4. Integrating with respect to $r$ and $\theta$ Now we multiply by the Jacobian $r$ and integrate over the radial and angular bounds. $V = \int_{0}^{2\pi} d\theta \int_{0}^{\sqrt{2}} (4r - 2r^3) \, dr$ Separating the angular part (which yields $2\pi$) and performing the power rule on $r$: $\int_{0}^{\sqrt{2}} (4r - 2r^3) \, dr = \left[ 2r^2 - \frac{1}{2}r^4 \right]_{0}^{\sqrt{2}}$ The antiderivative is evaluated at the boundary of the circular shadow in the $xy$-plane.

5. Final Computation Substitute the upper limit $r = \sqrt{2}$ into the evaluated integral. $2\pi \left( 2(\sqrt{2})^2 - \frac{1}{2}(\sqrt{2})^4 \right) = 2\pi (4 - 2) = 4\pi$ The final calculation combines the circular cross-section area and the height distribution to provide total volume.

Final Answer

The volume of the region $R$ is: $\boxed{4\pi}$

Common Mistakes

• Forgetting the Jacobian: A frequent error is forgetting to include the factor of $r$ when converting $dx\,dy$ to $dr\,d\theta$ in cylindrical/polar coordinates.
• Incorrect Bounds: Students often use the "4" from the $z$-intercept as the radius squared; it is vital to solve for the intersection $x^2 + y^2 = 4 - x^2 - y^2$ to find $r^2 = 2$.

Related Topics: Triple Integrals in Cylindrical Coordinates, Quadric Surfaces, Volume of Solids of Revolution.