Question

Show that the volume of the solid formed by the surface \(z=x^{a} y^{b}\), the \(x y-, y z-\), and \(x z\) -planes, and the plane parallel to the \(z\) -axis and going through the points \(\left(0, y_{0}\right)\) and \(\left(x_{0}, 0\right)\) is $$ \frac{x_{0}^{a+1} y_{0}^{b+1}}{a+b+2} B(a+1, b+1) $$.

Short answer

The volume of the solid under the surface \(z = x^a y^b\) and between the xy-, yz-, and xz-planes, and a plane parallel to the z-axis through the points (0, \(y_0\)) and (\(x_0\), 0) is indeed \(\frac{x_{0}^{a+1} y_{0}^{b+1}}{a+b+2} B(a+1, b+1)\). This is confirmed by setting up and calculating the double integral representing the volume, and then applying properties of the Beta function to match the given formula.

Step-by-step solution

Understand the Problem

We are given a surface in three dimensions represented by the function \(z = x^a y^b\). We are also given a set of constraints forming a volume bounded by the given function and the xy-plane, yz-plane, and xz-plane.

Set Up the Volume Integral

The volume of a solid formed above a region in the xy-plane and beneath a surface \(z = f(x, y)\) is given by the double integral \(\int \int f(x, y) dA\). Here, \(f(x, y)\) is given by \(x^a y^b\), and the region is a rectangle with vertices at (0,0), (0, \(y_0\)), (\(x_0\), 0), and (\(x_0\), \(y_0\)). Therefore, the volume integral is from 0 to \(y_0\) along the y-axis, and from 0 to \(x_0\) along the x-axis: \(\int_0^{x_0} \int_0^{y_0} x^a y^b dy dx\).

Evaluate the Integral

First integrate with respect to y: \(\int_{0}^{y_{0}} y^{b} dy = \frac{y_{0}^{b + 1}}{b + 1}\). The double integral becomes: \(\int_0^{x_0} x^a \frac{y_0^{b + 1}}{b + 1} dx\). Next, integrate with respect to x: \(\int_0^{x_0} x^a dx = \frac{x_0^{a + 1}}{a + 1}\), which gives the volume as \(\frac{x_{0}^{a + 1} y_{0}^{b + 1}}{(a + 1)(b + 1)}\).

Apply Beta Function Identity

The given formula involves a term called the Beta function, \(B(x, y)\), which has the identity \(B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}\). Also, \(\Gamma(x) = (x - 1)!\) when \(x\) is a positive integer. Using these facts, one can show that \(B(a + 1, b + 1) = (a!) (b!) / (a + b + 1)! \). Therefore, \((a + 1) (b + 1) B(a + 1, b + 1) = a!b! / (a + b + 2)!\), and hence the final quoted volume result follows given the computed volume from the prior step.

Key concepts

Beta Function

The Beta Function, denoted by \( B(x, y) \), is a special function in mathematics, particularly within the realms of calculus and analysis. It's a symmetric function, meaning \( B(x, y) = B(y, x) \), and is closely related to the Gamma Function. The Beta Function is defined by the integral: \[B(x, y) = \int_0^1 t^{x-1} (1-t)^{y-1} \, dt\]This function plays a crucial role in problems involving products of powers of variables within integrals. It shows up in probability and statistics, helping calculate operating intervals or intervals of deviation. In our original problem, the identity \( B(a+1, b+1) = \frac{a!b!}{(a+b+2)!} \) is used to find the volume of a certain three-dimensional region formed by bounded surfaces.

• It can simplify complex integral calculations when evaluating volumes under surfaces.
• Works with gamma function properties to bridge discrete and continuous cases.

Double Integral

In calculus, the Double Integral is an extension of the basic concept of integration applied to functions of two variables. It is used to compute the volume under a surface \( z = f(x,y) \) across a specified region in the xy-plane. The double integral is expressed as: \[\int\int_R f(x, y) \, dA\]This means you are integrating a function over a two-dimensional space or region \( R \). For the exercise, the function \( z = x^a y^b \) dictates the height of the surface over the rectangle \([0, x_0] \times [0, y_0]\). The steps involve:

• Integrating first with respect to one variable while keeping the other constant.
• Then, integrating the resulting expression with respect to the second variable.

By using a double integral, you can find volumes, center of mass, and surface areas in three-dimensional spaces. It greatly simplifies multivariable integration, especially when dealing with complex domains.

Three-Dimensional Calculus

Three-Dimensional Calculus expands the principles of single-variable calculus into three dimensions, handling variables and functions in a space described by \( x, y, \) and \( z \). It is fundamental for exploring volumes and areas in spaces beyond just flat planes. In our problem, we use three-dimensional calculus to solve for the volume formed by desires \( z = x^a y^b \) and constrained by the planes \( xy, yz, \) and \( xz \).

• It also involves concepts like divergence, curl, and gradient in vector calculus.
• Utilizes cylindrical or spherical coordinates for solving problems with circular or spherical symmetries.

Applications range from traditional physics questions to advanced topics in engineering and computer graphics. Understanding three-dimensional calculus enhances comprehension of movement, flow, and distribution in physical spaces.

Gamma Function

The Gamma Function, \( \Gamma(n) \), is a generalization of the factorial function \( n! \) to non-integer and complex numbers. For positive integers, it satisfies \( \Gamma(n) = (n-1)! \). It is defined by the integral: \[\Gamma(n) = \int_0^\infty t^{n-1} e^{-t} \, dt\]When interacting with the Beta Function in calculus problems, the Gamma Function helps in simplifying factorial terms into continuous domains. It allows for the smooth handling of limits and making calculations more manageable in complex integrals. In our solved problem, understanding that \( \Gamma(x) = (x-1)! \) helps relate the Beta Function, making it easier to factor out terms and compute the desired volume.

• Useful in statistical distributions, such as the normal and gamma distributions.
• Provides solutions to integral equations common in physics and engineering.

Mastering the Gamma Function allows for tackling a broad range of problems in mathematical analysis, especially when dealing with growth and scaling behaviors.