Problem 12 - APEX Calculus

Question

(a) Sketch the region \(R\) given by the problem. (b) Set up the iterated integrals, in both orders, that evaluate the given double integral for the described region \(R\) (c) Evaluate one of the iterated integrals to find the signed volume under the surface \(z=f(x, y)\) over the region \(R .\) \(\iint_{R} x^{2} y d A,\) where \(R\) is bounded by \(y=\sqrt[3]{x}\) and \(y=x^{3} .\)

Step-by-Step Solution

Answer

The signed volume under the surface is \( \frac{8}{99} \).

Step 1: Sketch the Region R

To sketch the region \( R \), firstly plot the curves \( y = \sqrt[3]{x} \) and \( y = x^3 \). Determine their intersections to find the bounds of \( R \). Since \( y = \sqrt[3]{x} \) is the cube root function, it mirrors about the line \( y = x \) with points such as (0,0) and (1,1). The curve \( y = x^3 \) is a cubic function also passing through (0,0) and (1,1). Both curves intersect at (0,0) and (1,1). The region \( R \) is the area between these curves from \( x = 0 \) to \( x = 1 \).

Step 2: Set up the Iterated Integral with xy-order

Integrate first with respect to \( y \), then \( x \). The lower bound for \( y \) is the curve \( y = x^3 \), and the upper bound is the curve \( y = \sqrt[3]{x} \). The limits for \( x \) are from 0 to 1.\[\int_{0}^{1} \int_{x^3}^{\sqrt[3]{x}} x^2 y \, dy \, dx\]

Step 3: Set up the Iterated Integral with yx-order

Integrate first with respect to \( x \), then \( y \). The curve \( x = y^3 \) is the inversion of \( y = x^{1/3} \), and the curve \( x = y^{1/3} \) is the inversion of \( y = x^3 \). For fixed \( y \), \( x \) ranges from \( y^3 \) to \( y^{1/3} \). For \( y \), limits go from 0 to 1.\[\int_{0}^{1} \int_{y^3}^{y^{1/3}} x^2 y \, dx \, dy\]

Step 4: Evaluate the Iterated Integral with xy-order

Choose the xy-order integral \( \int_{0}^{1} \int_{x^3}^{\sqrt[3]{x}} x^2 y \, dy \, dx \). Start by evaluating the inner integral:\[\int_{x^3}^{\sqrt[3]{x}} x^2 y \, dy = x^2 \left[ \frac{y^2}{2} \right]_{x^3}^{\sqrt[3]{x}} = x^2 \left[ \frac{(\sqrt[3]{x})^2}{2} - \frac{(x^3)^2}{2} \right]\]Simplify this to obtain:\[\frac{x^2 (x^{2/3})}{2} - \frac{x^2 (x^6)}{2} = \frac{x^{8/3} - x^8}{2}\]Now, integrate with respect to \( x \):\[\int_{0}^{1} \frac{x^{8/3} - x^8}{2} \, dx = \frac{1}{2} \left[ \frac{x^{11/3}}{11/3} - \frac{x^9}{9} \right]_0^1\]\[= \frac{1}{2} \left( \frac{3}{11} - \frac{1}{9} \right) = \frac{1}{2} \left( \frac{27}{99} - \frac{11}{99} \right) = \frac{1}{2} \times \frac{16}{99} = \frac{8}{99}\]

Key Concepts

Iterated Integrals

Iterated integrals play an essential role in the evaluation of double integrals. They allow us to integrate a function over a two-dimensional region by breaking it down into simpler one-dimensional integrals. In the context of the exercise, we have a bivariate function that depends on two variables \( x \) and \( y \).
To set up an iterated integral, we need to choose the order of integration. The order can either be \( xy \) (integrate first with respect to \( y \), then \( x \)) or \( yx \) (integrate first with respect to \( x \), then \( y \)). Each order requires appropriate limits that describe the region of integration. In this particular case, the region is bounded by the curves \( y = \sqrt[3]{x} \) and \( y = x^3 \), with intersections at points (0,0) and (1,1).

• For \( xy \)-order: \( \int_{0}^{1} \int_{x^3}^{\sqrt[3]{x}} x^2 y \; dy \; dx \).
• For \( yx \)-order: \( \int_{0}^{1} \int_{y^3}^{y^{1/3}} x^2 y \; dx \; dy \).

Using iterated integrals efficiently calculates the area or volume related to the function over the specified region.

Volume Under a Surface

The concept of volume under a surface is extended from finding areas under curves in single-variable calculus to finding signed volumes in multivariable calculus using double integrals. In this exercise, we consider the bivariate function \( f(x, y) = x^2 y \). The double integral of this function over the region \( R \) gives the signed volume between the surface \( z = f(x, y) \) and the \( xy \)-plane.
When evaluating the volume under the surface, the sign of the integral indicates the net volume. A positive value indicates that the major part of the region lies above the \( xy \)-plane, while a negative value indicates it lies below.
In our solution, the integral evaluated to \( \frac{8}{99} \), indicating a positive volume, meaning the region above the \( xy \)-plane is dominant in this configuration.

Definite Integration

Definite integration extends the concept of summing over a specific interval to multivariable functions. It is pivotal in calculating both areas and volumes. In double integrals, definite integration applies twice, once for each variable of the function.
The steps in definite integration involve choosing appropriate limits, which, in cases of iterated integrals, need careful determination from the region's boundaries. As seen, depending on the path of integration, we either integrate first with respect to \( y \) and then \( x \), or vice versa. These limits result from the intersection points of the bounding curves and dictate the precise range where the integral evaluates.
In the exercise, the limits \( 0 \) to \( 1 \) for \( x \) and between \( x^3 \) and \( \sqrt[3]{x} \) for \( y \), or equivalently \( y^3 \) to \( y^{1/3} \) for \( x \) with \( y \) from \( 0 \) to \( 1 \), were chosen for definite integration.

Bivariate Functions

Bivariate functions are functions of two independent variables, typically \( x \) and \( y \). These are essential when we want to describe surfaces in three-dimensional space. In the context of the original exercise, the function \( f(x, y) = x^2 y \) is a bivariate function.
Understanding these functions allows us to study complex surfaces and make sense of their shape and behavior over specific regions. In practical applications, bivariate functions can be used to model physical phenomena, optimize designs, or predict outcomes in engineering and science.

• Bivariate functions play a crucial role in evaluating double integrals, as they define the surface beneath which we calculate volumes or areas.
• These functions generalize the concept of one-dimensional integration into two dimensions, providing vast insight into spatial constructs.

Overall, grasping bivariate functions enables us to solve numerous practical problems using mathematical modeling.