Problem 32 - Advanced Engineering Mathematics

Question

Find the volume of the solid bounded by the graphs of the given equations. $$ z=4-x^{2}-\frac{1}{4} y^{2}, z=0 $$

Step-by-Step Solution

Answer

The volume of the solid is 32 units cubed.

Step 1: Understanding the Problem

We need to find the volume of the solid bounded by two surfaces: the surface described by the equation \(z = 4 - x^2 - \frac{1}{4} y^2\) and the surface \(z = 0\). The equation \(z = 4 - x^2 - \frac{1}{4} y^2\) represents a paraboloid opening downward, and \(z = 0\) represents the xy-plane. The region of interest is enclosed between these two surfaces.

Step 2: Identify the Region of Integration

To find where these two surfaces intersect, we set \(z = 0\) in the equation of the paraboloid: \[0 = 4 - x^2 - \frac{1}{4} y^2.\]Rearranging gives: \[x^2 + \frac{1}{4} y^2 = 4.\]This is an ellipse in the xy-plane with semi-major axis 2 along the x-axis and semi-minor axis 4 along the y-axis.

Step 3: Set Up the Double Integral for Volume

The volume of the solid can be found using the double integral of the function that defines the height, \(z = 4 - x^2 - \frac{1}{4} y^2\), over the region defined by the ellipse. The integral is: \[V = \int_{y=-4}^{4} \int_{x=-2\sqrt{1-(\frac{y^2}{16})}}^{2\sqrt{1-(\frac{y^2}{16})}} (4 - x^2 - \frac{1}{4} y^2) \; dx \, dy.\]

Step 4: Solve the Inner Integral

First, integrate with respect to \(x\):\[\int_{x=-2\sqrt{1-(\frac{y^2}{16})}}^{2\sqrt{1-(\frac{y^2}{16})}} (4 - x^2 - \frac{1}{4} y^2) \, dx.\]This gives:\[4x - \frac{x^3}{3} - \frac{1}{4} y^2 x \Bigg|_{-2\sqrt{1-(\frac{y^2}{16})}}^{2\sqrt{1-(\frac{y^2}{16})}}.\]

Step 5: Solve the Outer Integral

After solving the inner integral, compute the difference and then integrate with respect to \(y\) over the interval from \(-4\) to \(4\).\[\int_{-4}^{4} g(y) \, dy,\]where \(g(y)\) is the integral result from step 4. This step requires evaluating the resulting function, simplifying and integrating over \([ -4, 4 ]\).

Step 6: Compute the Final Volume

Once the integration over \(y\) is complete, calculate the definite integral value to find the final volume. Adjust any constants if needed and ensure the result is positive, as volume can't be negative.

Key Concepts

Double Integral

When calculating the volume of a solid bounded by curved surfaces, mathematicians often use a technique called double integration. Imagine you want to find the volume under a surface in a certain region on the xy-plane. Instead of calculating one area at a time, we perform a double integral. This allows us to consider infinitesimally small changes across both the x and y directions simultaneously.
Here’s how it works:

• The first integral is taken with respect to one variable, such as x, over a specific range.
• The second integral is then taken with respect to the other variable, y, again over a specified range.

Both integrals together allow us to sweep across the plane and right below the surface, effectively summing up the tiny volumes. This aggregated sum provides the total volume of the solid under consideration. For our problem with the paraboloid, the double integral collects bits of area under the paraboloid, calculating the volume below the surface described by the function hauteur \((z = 4 - x^2 - \frac{1}{4} y^2)\). This way, all parts of the solid are considered and combined efficiently.

Ellipse in xy-plane

An ellipse in the xy-plane is a curve that looks like a stretched circle. In our example, when setting the paraboloid's height to zero \((z=0)\), the equation becomes \(x^2 + \frac{1}{4} y^2 = 4\). This describes an ellipse.
Different from a circle that has the same radius in all directions, an ellipse has a major axis and a minor axis:

• The semi-major axis is along the x-axis, measuring 2 units. This means the widest part of the ellipse goes horizontally.
• The semi-minor axis is parallel to the y-axis, measuring 4 units. Thus, the ellipse stretches further vertically.

Understanding the ellipse's shape is crucial because it defines the region for our double integral. The x and y boundaries need to fit within this ellipse to accurately compute the volume. In terms of integration, using these ellipse properties helps set correct limits for x and y.

Volume of Solids

Calculating the volume of a solid requires understanding its boundaries and the space it fills. We use math to describe these surfaces and their behavior in space. In this particular problem, our solid is bounded by a paraboloid and the xy-plane.
To determine volume:

• Identify the equation for the 3D shape (here, the paraboloid \(z = 4 - x^2 - \frac{1}{4} y^2\)).
• Locate the region in the xy-plane that the solid projects onto, which is established as an ellipse.
• Perform a double integration over the determined region, calculating the sum of all tiny volumes under the surface from \(z=0\) up to the paraboloid surface.

The result of this integration provides a numerical value representing the volume, ensuring it accounts for every part of the space the solid occupies under the paraboloid and above the xy-plane.

Surface of Revolution

A surface of revolution appears when a curve revolves around an axis, generating a 3D object. Imagine turning a simple 2D arc or line around an axis of rotation, like spinning a wheel, which creates a symmetric surface in three-dimensional space.
For our paraboloid problem, while it's not strictly a surface of revolution by itself, understanding these can help visualize similar surfaces. When a parabolic slice rotates around an axis like the z-axis, it forms a paraboloid.
The symmetry and shape insights from such surfaces assist in calculating volumes because:

• These shapes often lead to simplified calculations, taking advantage of their symmetry.
• Integrals are easier to set up with known boundary behaviors, reducing errors.

While the given exercise involves static integration techniques rather than dynamic rotation, the understanding of symmetrical 3D forms like paraboloids alongside these revolutions provides foundational insight into diverse volumetric computations.