Problem 25 - University Calculus: Early Transcendentals

Question

Find the volume of the region bounded above by the paraboloid \(z=x^{2}+y^{2} \quad\) and below by the square \(\quad R:-1 \leq x \leq 1\) \(-1 \leq y \leq 1\)

Step-by-Step Solution

Answer

The volume is \(\frac{8}{3}\).

Step 1: Understand the region defined by the paraboloid

The equation of the paraboloid is given by \( z = x^2 + y^2 \). This surface opens upwards with the vertex at the origin (0, 0, 0). The region of interest is above this paraboloid, up to the xy-plane (\(z=0\)).

Step 2: Define the limits of integration based on the square

The region in the xy-plane is defined as \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\). These are the limits for the integration over x and y.

Step 3: Set up the double integral for the volume

The volume of the region between the paraboloid and the xy-plane can be found using the double integral: \[ \int_{-1}^{1} \int_{-1}^{1} (x^2 + y^2) \, dy \, dx \] Here, the function to integrate is \(x^2 + y^2\). The inner integral is with respect to \(y\) and the outer integral is with respect to \(x\).

Step 4: Integrate with respect to y

Compute the first integral with respect to y:\[ \int_{-1}^{1} (x^2 + y^2) \, dy = \int_{-1}^{1} (x^2 \, dy) + \int_{-1}^{1} (y^2 \, dy) \].This results in:\[ [x^2y + \frac{y^3}{3}]_{-1}^{1} \].Evaluating at the bounds gives:\[ x^2(1) + \frac{1}{3}(1) - (x^2(-1) + \frac{(-1)^3}{3}) = 2x^2 + \frac{2}{3} \].

Step 5: Integrate with respect to x

Now, integrate the result with respect to x:\[ \int_{-1}^{1} (2x^2 + \frac{2}{3}) \, dx \].Calculate:\[ 2 \int_{-1}^{1} x^2 \, dx + \frac{2}{3} \int_{-1}^{1} dx \],which results in:\[ 2 \cdot \frac{x^3}{3} \Big|_{-1}^{1} + \frac{2}{3} \cdot x \Big|_{-1}^{1} \].Solving these gives:\[ \left(2 \cdot \frac{1}{3} - 2 \cdot \frac{-1}{3}\right) + \left(\frac{2}{3} \cdot (1 - (-1))\right) = 2 \cdot \frac{2}{3} + \frac{4}{3} = \frac{8}{3} \].

Step 6: Conclusion

Therefore, the volume of the region bounded by the paraboloid \( z = x^2 + y^2 \) and the square region in the xy-plane is \( \frac{8}{3} \).

Key Concepts

Double Integration

Double integration is an extension of single-variable integration into higher dimensions. It allows us to calculate volumes and areas in three-dimensional space by summing up infinitesimal pieces. In our problem, we are using double integration to find the volume enclosed below a paraboloid and above a portion of the xy-plane.

The process involves two successive integrations. The inner integral computes the area under the curve in the y-direction, holding x constant. Then, the outer integral sums these "slices" across the x-direction to find the total volume. Specifically, for the given problem, our double integral is set as:

• Inner integral: \[ \int_{-1}^{1} (x^2 + y^2) \, dy \] which evaluates the slice for a given x.
• Outer integral: \[ \int_{-1}^{1} \text{{(result of inner integral)}} \, dx \]

This structured approach helps systematically address the complexity that comes with evaluating the area and volume enclosed by three-dimensional surfaces.

Paraboloid

A paraboloid is a three-dimensional surface with a parabolic cross-section in at least one direction. In the exercise we're discussing, the paraboloid is represented by the equation \( z = x^2 + y^2 \). This equation shows that for any fixed values of x and y, z will be non-negative and increase as we move away from the origin (0,0,0).

Visualizing this, imagine an upward-opening bowl shape, symmetrically extending around the z-axis. The further from the origin you move on the xy-plane, the higher the surface of the paraboloid goes. Here are key characteristics of our specific paraboloid:

• Vertex of the paraboloid is at the origin \((0, 0, 0)\).
• It is symmetric about the z-axis, meaning any rotation around the z-axis doesn't change its shape.
• The cross-sections parallel to the xy-plane (horizontal planes) are circular.

Understanding these properties helps in visualizing how the integration limits affect the region over which the volume is computed.

Limits of Integration

Limits of integration are fundamental parameters in setting up integrals, defining the region over which the integral is evaluated. In our volume calculation problem, these limits are determined by the boundary in the xy-plane and the surface of the paraboloid.

The specific boundaries for x and y are given as \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\). These form a square in the xy-plane over which we integrate. The limits can be visualized as follows:

• The integration in the y-direction goes from -1 to 1 for each fixed x, covering the full height of the square within each slice.
• For the x-direction, the integration also spans from -1 to 1, summing all the vertical slices across the width of the square.

The paraboloid provides the upper limit in the z-direction for each point within these x and y boundaries. Properly understanding and setting these limits ensures that the double integral accurately computes the volume beneath the paraboloid relative to the xy-plane block defined.