Problem 16 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral evaluates to approximately 2.41.

Step 1: Understand the Surface

The surface of integration is defined by the equation \( z = x^2 + y \), with bounds \( 0 \leq x \leq 1 \) and \( -1 \leq y \leq 1 \). This is a parabolic surface stretched over a rectangular domain in the \( xy \)-plane.

Step 2: Set Up the Surface Integral

The surface integral of a function \( G(x, y, z) \) over a given surface \( S \) is given by \( \iint_{S} G(x, y, z) \, dS \). Here, \( G(x, y, z) = x \).

Step 3: Calculate the Differential Element of Area

To find \( dS \), the differential area element of the surface, we use the parameterization of the surface \( \vec{r}(x, y) = (x, y, x^2+y) \). The partial derivatives are \( \vec{r}_x = (1, 0, 2x) \) and \( \vec{r}_y = (0, 1, 1) \). The cross-product \( \vec{r}_x \times \vec{r}_y = (-2x, -1, 1) \), and consequently, \( ||\vec{r}_x \times \vec{r}_y|| = \sqrt{4x^2 + 2} \). Thus, \( dS = \sqrt{4x^2 + 2} \, dx \, dy \).

Step 4: Setup the Double Integral

The surface integral becomes: \[ \iint_{0 \leq x \leq 1, \; -1 \leq y \leq 1} x \sqrt{4x^2 + 2} \, dy \, dx. \]

Step 5: Integrate with respect to y

The integrand \( x \sqrt{4x^2 + 2} \) is independent of \( y \), so the integral with respect to \( y \) is simply \( \int_{-1}^{1} dy = 2 \). The integral reduces to: \[ \int_{0}^{1} 2x \sqrt{4x^2 + 2} \ dx. \]

Step 6: Integrate with respect to x

To solve \( \int_{0}^{1} 2x \sqrt{4x^2 + 2} dx \), perform a substitution: let \( u = 4x^2 + 2 \), then \( du = 8x \, dx \). Consequently, \( dx = \frac{du}{8x} \), and splitting yields \( x \sqrt{u} \cdot \frac{du}{8x} = \frac{\sqrt{u}}{8} \, du \). Adjust limits accordingly: when \( x = 0, u = 2 \), and when \( x = 1, u = 6 \). So the integral becomes \( \frac{1}{8} \int_{2}^{6} \sqrt{u} \, du \).

Step 7: Solve the Substituted Integral

The integral \( \frac{1}{8} \int_{2}^{6} u^{1/2} \, du = \frac{1}{8} \left[ \frac{2}{3} u^{3/2} \right]_2^6 \). Calculating gives: \[ = \frac{1}{8} \left( \frac{2}{3} (6^{3/2}) - \frac{2}{3} (2^{3/2}) \right) \]Simplifying further: \( 6^{3/2} = 6 \sqrt{6} \) and \( 2^{3/2} = 2 \sqrt{2} \); the expression becomes \[ \frac{1}{8} \left( \frac{2}{3} (6 \sqrt{6} - 2 \sqrt{2}) \right) = \frac{1}{12} (6 \sqrt{6} - 2 \sqrt{2}). \]

Step 8: Calculate the Final Result

Finally compute \( \frac{1}{12} (6 \sqrt{6} - 2 \sqrt{2}) \). Numerically, this evaluates to approximately 2.41.

Key Concepts

Parabolic Surface

A parabolic surface in mathematics is formed when a parabola is extended along a plane. In our case, this surface is characterized by the equation \( z = x^2 + y \). This represents a paraboloid, except instead of curving downward or upward in the \( z \)-direction alone, it is influenced by both the \( x \) and \( y \)-axes.

Here, when \( x \) varies between 0 and 1 and \( y \) ranges from -1 to 1, we create a surface over a rectangular domain in the \( xy \)-plane.
This shape dictates that the surface curves gently following the square of the \( x \)-value plus the \( y \)-value at each point.

• If we were to "slice" this surface along a vertical line through \( x \), we would see a parabolic curve.
• A similar slice along \( y \) would display linear behaviors, as only \( x \) contributes to the parabolic shape.

Understanding this helps visualize how the surface changes as \( x \) or \( y \) changes, and is pivotal when setting up integrations.

Differential Area Element

In surface integrals, the differential area element \( dS \) plays a crucial role. It represents a tiny "patch" of surface area over which you integrate the function.
For the parabolic surface \( z = x^2 + y \), we use parameterization, defining the surface in terms of two variables \( x \) and \( y \):
\( \vec{r}(x, y) = (x, y, x^2 + y) \).

• We first calculate partial derivatives: \( \vec{r}_x = (1, 0, 2x) \) and \( \vec{r}_y = (0, 1, 1) \).
• The differential area element is found by taking the cross product of these derivatives: \( \vec{r}_x \times \vec{r}_y = (-2x, -1, 1) \).
• The magnitude of this cross product gives us \( dS = \sqrt{4x^2 + 2} \, dx \, dy \).

In essence, \( dS \) quantifies how the surface "spreads out," responding to changes in the direction of both \( x \) and \( y \). This is useful for computing how functions stretch over a surface.

Double Integral

Double integrals extend the idea of a single integral into two dimensions. They're essential in calculating the total sum or accumulation over an area, or in this case, over a surface.
For the function \( G(x, y, z) = x \) over the specified surface, we express the surface integral as a double integral:
\[ \iint \limits_{0 \leq x \leq 1, \; -1 \leq y \leq 1} x \sqrt{4x^2 + 2} \, dy \, dx. \]

• The integrand \( x \sqrt{4x^2 + 2} \) results from multiplying \( x \) by the magnitude of the cross-product, reflecting \( dS \).
• Integration limits reflect the domain of \( x \) and \( y \), shaping the integrated area.

To compute this integral:- First, integrate with respect to \( y \). This is straightforward as \( x \sqrt{4x^2 + 2} \) is independent of \( y \).- After solving for \( y \), integrate the resulting function with respect to \( x \) to evaluate the total accumulation on the surface.

Substitution Method

The substitution method in integration helps simplify the integrand into a more manageable form. It mimics changes of variable in basic arithmetic. For our task, integrating \( \int_{0}^{1} 2x \sqrt{4x^2 + 2} \, dx \), substitution aids in tackling the square root.

• Select \( u = 4x^2 + 2 \) which simplifies the expression inside the square root.
• Next, calculate \( du = 8x \, dx \), leading to \( dx = \frac{du}{8x} \), which we substitute back into the integral.

The limits also change:

• When \( x = 0, u = 2 \).
• When \( x = 1, u = 6 \).

The integral becomes \( \frac{1}{8} \int_{2}^{6} \sqrt{u} \, du \), which is more straightforward to solve. This method simplifies complex chains, aiding in obtaining results that are otherwise intricate to approach with original variables.