Problem 47 - University Calculus: Early Transcendentals

Question

Show that the volume of the segment cut from the paraboloid $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{z}{c}$$ by the plane \(z=h\) equals half the segment's base times its altitude.

Step-by-Step Solution

Answer

The volume is half the product of the base area and the altitude.

Step 1: Identify the Region

Given the equation of the paraboloid \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{z}{c} \), we need to find the volume of the segment between \( z=0 \) (vertex of the paraboloid) and \( z=h \) (the plane that cuts the paraboloid).

Step 2: Express z in Terms of x and y

Rearrange the equation of the paraboloid to express \( z \) in terms of \( x \) and \( y \): \[ z = c \left( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \right). \]

Step 3: Set Up the Volume Integral

To find the volume, integrate the area over the circular region in the \( xy \)-plane. The limits of \( z \) are from \( 0 \) to \( h \), so consider:\[ V = \int_0^h A(z) \, dz, \] where \( A(z) \) is the area of the elliptical region at height \( z \).

Step 4: Find A(z), the Area of Ellipse

For a fixed \( z \), the equation \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{z}{c} \) describes an ellipse. Solve for the area of this ellipse, \[ A(z) = \pi a b \frac{z}{c}. \]

Step 5: Integrate to Find the Volume

Substitute \( A(z) \) into the integral:\[ V = \int_0^h \left( \pi a b \frac{z}{c} \right) \, dz. \]Integrate with respect to \( z \):\[ V = \pi a b \frac{1}{c} \left. \frac{z^2}{2} \right|_0^h = \frac{\pi a b}{2c} h^2. \]

Step 6: Relate to Half the Base Times Altitude

The base of the segment is the area of the ellipse at \( z = h \), which is \( \pi a b \frac{h}{c} \). The altitude is the height \( h \). Half the segment's base times its altitude is:\[ \frac{1}{2} \pi a b \frac{h}{c} \times h = \frac{\pi a b}{2c} h^2, \]which matches the expression for the volume.

Key Concepts

Paraboloid

A paraboloid is a three-dimensional surface generated by rotating a parabola around its axis of symmetry. It can take on two main shapes:

• Elliptical Paraboloid: This shape resembles an elongated dish and is defined by the equation \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{z}{c} \). The cross-sections perpendicular to its principal axis (z-axis in this case) are ellipses.
• Hyperbolic Paraboloid: This shape looks like a saddle and is defined by the equation \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = \frac{z}{c} \).

The problem involves an elliptical paraboloid. When sliced by a plane, this shape creates sections whose volumes can be calculated. Understanding the paraboloid can help in visualizing both the geometric and algebraic transformations involved in calculus volume integration.

Definite Integral

The concept of a definite integral is a cornerstone of calculus. It is used to find the exact area under a curve within a specified interval.

• In this context, we use definite integrals to calculate the volume of a solid of revolution or segments of geometric shapes.
• The notation for a definite integral is \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the boundaries of the interval and \( f(x) \) is the function under analysis.

For the paraboloid volume, the definite integral \( V = \int_0^h A(z) \, dz \) determines the accumulation of elliptical area slices from \( z = 0 \) to \( z = h \). This showcases how definite integrals apply to giving precise measures of volumes in three-dimensional space.

Volume of Solids

Calculating the volume of solids involves integrating cross-sectional areas of the shape. The integral of these areas with respect to height gives us the total volume.

• A good starting point is expressing the solid's geometric properties through mathematical equations, such as \( z = c( \frac{x^2}{a^2} + \frac{y^2}{b^2}) \).
• The slice at any height \( z \) through this paraboloid is an ellipse, with area \( A(z) = \pi a b \frac{z}{c} \).
• By integrating these elliptical segments from the base to the top of the solid, using \( V = \int_0^h A(z) \, dz \), we derive the volume as \( \frac{\pi a b}{2c} h^2 \).

This expression connects the abstract calculus operation of integration with physical volume, simplifying complex surfaces into manageable measures.

Ellipse Area

The ellipse is a fundamental geometric figure with applications in multiple fields, including volume calculation in calculus.

• The equation for an ellipse in standard form is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the semi-major and semi-minor axes.
• The area of an ellipse is given by \( A = \pi a b \). This formula is crucial when finding slices of shapes like the paraboloid.
• In our specific problem, the formula \( A(z) = \pi a b \frac{z}{c} \) accounts for the changing size of the ellipse as it moves along the axis of the paraboloid at different heights \( z \).

Understanding how to calculate the area for different sections helps in forming the integral setups for finding volumes in calculus problems.