Problem 300 - Calculus Volume 3

Question

Evaluate \(\int_{S} x y z d S\) if \(S\) is the part of plane \(z=x+y\) that lies over the triangular region in the \(x y\) -plane with vertices (0,0,0),(1,0,0) , and (0,2,0) .

Step-by-Step Solution

Answer

The integral evaluates to \(\sqrt{3}\).

Step 1: Parametrize the Surface

First, recognize that the surface, given as the plane \(z = x + y\), can be parametrized over the triangular region in the \(xy\)-plane. Let it be defined as \(\vec{r}(x,y) = \langle x, y, x+y \rangle\). The region over which this is defined can be expressed as \(0 \leq x \leq 1\) and \(0 \leq y \leq 2 - 2x\).

Step 2: Calculate the Surface Element

To calculate the surface integral, you need the magnitude of the cross product of the partial derivatives of \(\vec{r}(x,y)\) with respect to \(x\) and \(y\). Calculate these as: \(\vec{r}_x = \langle 1, 0, 1 \rangle\) and \(\vec{r}_y = \langle 0, 1, 1 \rangle\). The cross product is \(\vec{r}_x \times \vec{r}_y = \langle -1, -1, 1 \rangle\). The magnitude is \(\sqrt{3}\). Therefore, the surface element \(dS = \sqrt{3}\, dx\, dy\).

Step 3: Formulate the Integral

Substitute the surface parametrization into the integrand: \(xy(x+y)\). Thus, the integral becomes \(\int_0^1 \int_0^{2-2x} xy(x+y) \sqrt{3}\, dy\, dx\).

Step 4: Evaluate the Inner Integral

Consider the inner integral \(\int_0^{2-2x} xy(x+y) \, dy\). Expand the integrand to get \(xy^2 + xy^2\). Compute the integral as \((x/3)y^3 + (xy/2)y^2\) and evaluate it from \(0\) to \(2-2x\).

Step 5: Evaluate the Outer Integral

Now perform the integration over \(x\). Substitute the result from the inner integral into \(\int_0^1 (...) \sqrt{3} \, dx\). Solve this integral to obtain the final result.

Key Concepts

Parametrization

When dealing with surface integrals, parametrization is an essential concept. It involves expressing the given surface in terms of two parameters, often denoted as variables like x and y. This converts a surface problem into a simpler problem on a plane. In our exercise, we work with a plane defined by the equation z = x + y. By parametrizing this surface, we map out the plane using a vector function:

• \[ \vec{r}(x,y) = \langle x, y, x + y \rangle \]

This vector function describes every point on the surface in terms of x and y, essentially transforming the original surface in three dimensions into two dimensions. This method not only simplifies calculations but also provides a clearer understanding of how the surface behaves across the specified region.

Cross Product

The cross product is crucial in evaluating surface integrals because it helps determine the orientation and area of the infinitesimally small surface elements. For the given plane in the exercise, we calculate the partial derivatives of our parametrization\(\vec{r}(x,y)\), which are as follows:

• \( \vec{r}_x = \langle 1, 0, 1 \rangle \)
• \( \vec{r}_y = \langle 0, 1, 1 \rangle \)

The cross product of these derivatives is found by:

• \( \vec{r}_x imes \vec{r}_y = \langle -1, -1, 1 \rangle \)

This vector is normal to our surface and the magnitude of this cross product, \( \sqrt{3} \), represents the area scaling for the surface element \( dS \). The surface element associated with our parametrized surface is therefore \( \sqrt{3} \, dx \, dy \), crucial for calculating the surface integral.

Plane Equation

The plane equation serves as the backbone of our surface description in multiple variable calculus. Here, the plane is given by the equation \( z = x + y \). Planes in mathematical problems can be visualized as flat, two-dimensional surfaces extending infinitely in 3D space. Specifically, they can be defined by a linear equation in terms of the three coordinates x, y, and z.
By rearranging, we express z in terms of x and y. Since we are interested in evaluating the integral over a specific region of this plane, this equation helps us set boundaries for our calculations by fixing z as a function of other parameters.
Understanding the equation of the plane allows us to parametrize the given surface, which is an integral step in evaluating a surface integral effectively and accurately.

Region in the xy-plane

The region in the xy-plane is foundational since it defines where our surface integral calculation takes place. In our exercise, this region is a triangular section. The vertices of the triangle are crucial in forming its boundaries: (0,0), (1,0), and (0,2).
These vertices involve inequalities for x and y:

• For x: \( 0 \leq x \leq 1 \)
• For y: \( 0 \leq y \leq 2-2x \)

These inequalities describe a triangular region within which the surface part exists, and they provide limits for integration. Understanding this region helps us accurately translate the behavior of the surface and ensure our integral captures only the specified part of the plane. This accurate mapping is critical in solving integrals over surfaces and ensuring the boundaries are correctly applied during calculations.