Problem 16 - Thomas Calculus

Question

Integrate \(G(x, y, z)=x\) over the surface given by $$z=x^{2}+y \quad \text {for} \quad 0 \leq x \leq 1, \quad-1 \leq y \leq 1$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{6} \left(6\sqrt{6} - 2\sqrt{2}\right) \).

Step 1: Understand the Problem

We need to integrate the function \( G(x, y, z) = x \) over the surface defined by \( z = x^2 + y \) where \( 0 \leq x \leq 1 \) and \( -1 \leq y \leq 1 \). This requires finding the surface integral \( \iint_S G(x, y, z) \, dS \) where the surface \( S \) is given implicitly by the equation \( z = x^2 + y \).

Step 2: Set Up the Surface Integral

The surface integral \( \iint_S G(x, y, z) \, dS \) will involve finding \( dS \), which is determined by the surface's parametrization. For a surface given by \( z = f(x, y) \), the differential area element is \( dS = \sqrt{1 + (f_x)^2 + (f_y)^2} \, dx \, dy \).

Step 3: Calculate Partial Derivatives

Compute the partial derivatives of \( f(x,y) = x^2 + y \). We have \( f_x = \frac{\partial}{\partial x}(x^2 + y) = 2x \) and \( f_y = \frac{\partial}{\partial y}(x^2 + y) = 1 \).

Step 4: Compute the Differential Area Element

Substitute the partial derivatives into the formula for \( dS \):\[ dS = \sqrt{1 + (2x)^2 + (1)^2} \, dx \, dy = \sqrt{1 + 4x^2 + 1} \, dx \, dy = \sqrt{2 + 4x^2} \, dx \, dy. \]

Step 5: Set Up the Integral

The function to integrate is \( G(x, y, z) = x \) over the region \( 0 \leq x \leq 1 \), \( -1 \leq y \leq 1 \). So, we set up the integral:\[ \iint_S x \, dS = \int_{x=0}^{1} \int_{y=-1}^{1} x \sqrt{2 + 4x^2} \, dy \, dx. \]

Step 6: Integrate with Respect to y

Since there is no \( y \) in the integrand, integrate with respect to \( y \) first:\[ \int_{-1}^{1} x \sqrt{2 + 4x^2} \, dy = \left[ 2x \sqrt{2 + 4x^2} \right]. \]

Step 7: Integrate with Respect to x

Now, integrate with respect to \( x \):\[ \int_{0}^{1} 2x \sqrt{2 + 4x^2} \, dx. \]Use substitution \( u = 2 + 4x^2 \), \( du = 8x \, dx \), switch limits accordingly to integrate this last step.

Step 8: Evaluate the Final Integral

Evaluating the integral with substitution:- When \( x=0 \), \( u = 2 \).- When \( x=1 \), \( u = 6 \).So integral is:\[ \frac{1}{4} \int_{2}^{6} \sqrt{u} \, du = \frac{1}{4} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{2}^{6} = \frac{1}{6} \left[ 6^{3/2} - 2^{3/2} \right]. \]Calculating this gives \( \frac{1}{6} \left[ 6\sqrt{6} - 2\sqrt{2} \right]. \)

Key Concepts

Multivariable Calculus

Multivariable calculus deals with functions of multiple variables and encompasses different techniques to evaluate derivatives and integrals in higher dimensions. Unlike single-variable calculus, it allows us to explore the behavior of systems in more than one dimension, which is essential since most real-world phenomena involve multiple factors. In this exercise, we're working with a function, \( G(x, y, z) = x \), over a surface defined by multiple variables, where our focus is on integrating over the entire specified region. This practice helps understand how changes in multiple dimensions can affect overall calculations, making multivariable calculus immensely important in fields ranging from engineering to economics.

Partial Derivatives

Partial derivatives are a cornerstone of multivariable calculus. They represent the rate of change of a function with respect to one variable while keeping others constant. In our context, for the surface given by \( z = f(x, y) = x^2 + y \), we need to compute the partial derivatives \( f_x \) and \( f_y \). This involves simply taking derivatives with respect to \( x \) and \( y \).
For \( f(x, y) = x^2 + y \):

• \( f_x = \frac{\partial}{\partial x}(x^2 + y) = 2x \)
• \( f_y = \frac{\partial}{\partial y}(x^2 + y) = 1 \)

These calculations are crucial because they feed directly into the determination of the differential area element, \( dS \), an essential component of calculating the surface integral.

Differential Area Element

The differential area element \( dS \) is a small piece of the surface area over which we're integrating. For surfaces defined as \( z = f(x, y) \), the formula for \( dS \) incorporates the partial derivatives \( f_x \) and \( f_y \) and gives us a way to translate the surface into an integrable area in the \( xy \)-plane.
Using the formula: \[ dS = \sqrt{1 + (f_x)^2 + (f_y)^2} \, dx \, dy \]
For our surface \( z = x^2 + y \):

• Substitute \( f_x = 2x \) and \( f_y = 1 \)
• \( dS = \sqrt{1 + (2x)^2 + (1)^2} \, dx \, dy = \sqrt{2 + 4x^2} \, dx \, dy \)

This gives us the differential element needed to properly account for every small patch of the surface, integral for determining the total integral over complex surfaces.

Integration Techniques

To evaluate the surface integral, we employ specific integration techniques including iterated integrals and substitution. First, ensure that the function within the integral can be separated into simpler components to make computation easier, starting with integrating with respect to \( y \): \[ \int_{-1}^{1} x \sqrt{2 + 4x^2} \, dy = \left[2x \sqrt{2 + 4x^2} \right] \] The next step involves integrating with respect to \( x \) and often requires substitution to simplify the integration process. Here, we perform a substitution where \( u = 2 + 4x^2 \) and \( du = 8x \, dx \).
This transforms the integral into a form that can be easily computed: \[ \frac{1}{4} \int_{2}^{6} \sqrt{u} \, du \] Evaluate this using fundamental integration rules and substitution bounds; such methods make complicated integrals more manageable and reveal the integral over a specific surface.