Problem 15 - Thomas Calculus

Question

Integrate \(G(x, y, z)=z-x\) over the portion of the graph of \(z=x+y^{2}\) above the triangle in the \(x y-\) plane having vertices \((0,\) \(0,0 ),(1,1,0),\) and \((0,1,0) .\) (See accompanying figure.)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{12} \).

Step 1: Understand the Integral Problem

The given problem requires integrating the function \( G(x, y, z) = z - x \) over the region where the surface \( z = x + y^2 \) lies above a triangular area in the \(xy\)-plane with vertices at \((0, 0), (1, 1), (0, 1)\).

Step 2: Set Up the Integration Region

The triangle in the \(xy\)-plane is bounded by the lines \(y = 0\), \(x = 0\), and \(y = x\). Hence, in this region, \(x\) varies from \(0\) to \(1\), and for a given \(x\), \(y\) varies from \(0\) to \(x\).

Step 3: Substitute z in G(x, y, z)

Since \( z = x + y^2 \) in the given surface, substitute \( z \) in the function: \( G(x, y, z) = (x + y^2) - x = y^2 \).

Step 4: Set Up the Integral

The integral of \( G(x, y, z) = y^2 \) over the triangular region is given by \[ \int_{0}^{1} \int_{0}^{x} y^2 \, dy \, dx. \]

Step 5: Integrate with Respect to y

Integrate \( y^2 \) with respect to \( y \) from \(0\) to \(x\): \[ \int_{0}^{x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{x} = \frac{x^3}{3}. \]

Step 6: Integrate with Respect to x

Now, integrate \( \frac{x^3}{3} \) with respect to \( x \) from \(0\) to \(1\): \[ \int_{0}^{1} \frac{x^3}{3} \, dx = \frac{1}{3} \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}. \]

Step 7: Conclusion: Solution to the Integral

The integral \( \int_{0}^{1} \int_{0}^{x} y^2 \, dy \, dx \), which represents the required integral of \( G(x, y, z) \) over the defined region, evaluates to \( \frac{1}{12} \).

Key Concepts

Integration over Regions

When dealing with integration over regions, you are working with functions defined over a specific area. This involves analyzing the extent of the region on which a function should be integrated. In our exercise, the region of interest is a triangular area in the xy-plane.
The vertices of this triangle are given by the coordinates (0, 0), (1, 1), and (0, 1). This specifies our bounds for integration. By identifying these points, we know the limits for the variables within the integration process.

• Understand the shape: A triangle determined by its vertices.
• Identify boundaries: Here, lines are y = 0, x = 0, and y = x.

With these, we can set up integrals properly by determining how x and y vary across the region. Specifically, x ranges from 0 to 1. For each fixed x, y varies from 0 up to x itself. This careful setup allows us to account for every point in the triangle when integrating the function.

Multivariable Calculus

Multivariable calculus extends single-variable calculus to functions of several variables. It is crucial in finding areas, volumes, and other applications in higher dimensions. In this exercise, we utilize double integrals to calculate over a defined surface.
Specifically, the function used is transformed based on the surface equation. Given the surface equation \(z = x + y^2\), we substitute to transform \(G(x, y, z)\) into a simpler function of x and y, which is \(y^2\).

• Function Transformations: Convert into simpler forms for integration.
• Double Integrals: Integrate iteratively over multiple variables.

By breaking down the integration process into steps: first, integrate with respect to y, and then with respect to x. This process combines calculus and logical setup, essential in evaluating integrals over surfaces and regions.

Surface Integration

Surface integration involves calculating integrals over surfaces, important in fields such as physics and engineering. Here, the surface is described by the function \(z = x + y^2\), which is a paraboloid.
Once the surface is defined, transforming the function for integration is necessary. We changed \(G(x, y, z) = z - x\) to \(y^2\) by substituting z from the surface equation.

• Surface Formation: Understand how the surface equation affects integration.
• Focus on the Region Above: Only integrate over specified sections of the surface.

Finally, perform the double integration over the region directly beneath the surface in the x-y plane. The solution \(\frac{1}{12}\) represents the accumulated function over the area of interest. Thus, surface integration provides insight into spatial functions' behavior over three-dimensional shapes.