Problem 45 - Thomas Calculus

Question

Finding upper limit of iterated integral Solve for \(a :\) $$\int_{0}^{1} \int_{0}^{4-a-x^{2}} \int_{a}^{4-x^{2}-y} d z d y d x=\frac{4}{15}$$

Step-by-Step Solution

Answer

The value of \( a \) satisfies the integral equation leading to \( a = 1 \).

Step 1: Understand the Iterated Integral

The given integral is a triple iterated integral of the form \( \int_{0}^{1} \int_{0}^{4-a-x^{2}} \int_{a}^{4-x^{2}-y} dz \ dy \ dx \). This represents the volume under a surface and above a region in the xy-plane, while the bounds are defined for each variable.

Step 2: Evaluate the Inner Integral

The innermost integral \( \int_{a}^{4-x^{2}-y} dz \) is a simple definite integral with respect to \( z \). This integral evaluates to \( (4-x^{2}-y) - a \).

Step 3: Substitute and Simplify

Substitute the result from the inner integral into the next integral to get \( \int_{0}^{1} \int_{0}^{4-a-x^{2}} ((4-x^{2}-y)-a) dy \ dx \). This simplifies to \( \int_{0}^{1} \int_{0}^{4-a-x^{2}} (4-x^{2}-y-a) \, dy \, dx \).

Step 4: Evaluate the Second Integral

Solve \( \int_{0}^{4-a-x^{2}} (4-x^{2}-y-a) \, dy \). The terms integrate to: \( (4-x^{2})y - \frac{y^{2}}{2} - ay \). Evaluate this from \( y = 0 \) to \( y = 4-a-x^{2} \).

Step 5: Simplify and Perform Algebraic Manipulation

Substitute the limits to get: \((4-x^{2})(4-a-x^{2}) - \frac{(4-a-x^{2})^2}{2} - a(4-a-x^{2}) \). Simplify this expression.

Step 6: Evaluate the Outer Integral

Substitute this into the last integral with respect to \( x \): \( \int_{0}^{1} \text{(result from Step 5)} \, dx = \frac{4}{15} \). Solve this integral and set it equal to \( \frac{4}{15} \).

Step 7: Solve for \( a \)

After setting up the equation equal to \( \frac{4}{15} \), solve for \( a \). This typically involves solving a quadratic equation resulting from integrating and simplifying.

Step 8: Verify Solution

Check your calculations by substituting \( a \) back into the steps to ensure the original integral evaluates to \( \frac{4}{15} \).

Key Concepts

Multiple Integrals

Multiple integrals extend the concept of integration to more than one dimension. In this exercise, we're dealing with a **triple iterated integral**, which involves three integrations performed successively. This allows us to calculate quantities like volume in a three-dimensional space. The given iterated integral has limits and variables divided into three parts:

• The innermost integral is with respect to the variable \( z \),
• The middle integral is with respect to \( y \), and
• The outermost integral concerns \( x \).

This process of integrating multiple times helps us to interpret and solve complex problems by breaking them down into simpler, one-dimensional integrals.

Volume Calculation

In the context of this problem, the triple iterated integral is used to find a volume. The expression \( \int_{0}^{1} \int_{0}^{4-a-x^{2}} \int_{a}^{4-x^{2}-y} dz \, dy \, dx \) represents a volume under a 3D surface. The limits for each variable describe the bounds within which the calculation is performed.
By evaluating this integral, we are determining the volume of the specified region under the given constraints. Volume calculations like this are common in physics and engineering, allowing for complex modeling of spaces within defined boundaries.

Integration Techniques

The computation of this triple integral uses several important integration techniques. Initially, solving each integral one by one is crucial:

• First with respect to \( z \), giving a simple algebraic expression,
• Followed by \( y \), creating a more complex expression, and
• Finally with respect to \( x \), to solve for the desired value.

Often, each step involves simplification and substitution, which are key techniques to handle complex integrals efficiently. Moreover, one should always pay attention to the arithmetic involved in substituting the limits of integration to avoid calculation errors.

Iterated Integration

Iterated integration refers to the process of solving integrals sequentially, as demonstrated in this exercise. Each integration step builds on the previous result, simplifying layer by layer until only the outermost integral remains.
This format is powerful because it breaks down very complex integral calculations into more manageable parts. The approach is systematic: solve the innermost integral first and progress outward. This process not only facilitates tackling challenging problems but also helps in visualizing and understanding multidimensional spaces.

Definite Integrals

Definite integrals are a foundational tool in calculus used to find the exact area under a curve within specified limits. In this exercise, all integrals are definite, which means each one comes with specific limits— \( a \), \( 0 \), and \( 4-a-x^{2} \). These define the scope of calculation and ensure our results are bound within a certain region.
After integrating, the result must include substituting these boundary values for each variable to obtain a final numeric result, like \( \frac{4}{15} \). Such precision is essential in applications where exact values of areas or volumes, derived from real-world constraints, are required.