Question

Find the area of the given surface over the region \(R\). \(f(x, y)=2 x+2 y+2 ; R\) is the triangle with corners (0,0) (1,0) and (0,1)

Short answer

The area of the surface is \(\frac{3}{2}\).

Step-by-step solution

Identify the Surface Equation

The surface equation given is \( f(x, y) = 2x + 2y + 2 \). This describes a plane in three-dimensional space.

Determine the Region of Integration

The region \( R \) is a triangular region with vertices at \((0,0), (1,0), \text{and } (0,1)\). This is a right-angled triangle in the xy-plane, with the right angle at the origin.

Set Up the Double Integral

The area of the surface over the region \( R \) is calculated using a double integral of the function over \( R \). Determine the bounds for integration. For \( y \), it ranges from 0 to \( 1-x \), and for \( x \), it ranges from 0 to 1.

Calculate the Jacobian

To find the area of the surface over the region, we need to multiply by the square root of \( 1 + (f_x)^2 + (f_y)^2 \), where \( f_x \) and \( f_y \) are partial derivatives. Calculate \( f_x = 2 \) and \( f_y = 2 \). The expression becomes \( \sqrt{1 + 2^2 + 2^2} = \sqrt{9} = 3 \).

Integrate to Find the Area

Set up the double integral: \( \int_{0}^{1} \int_{0}^{1-x} 3 \, dy \, dx \). First integrate with respect to \( y \): \( \int_{0}^{1-x} 3 \, dy = 3y \Big|_{0}^{1-x} = 3(1-x) \).

Evaluate the Integral

Substitute in the outer integral: \( \int_{0}^{1} 3(1-x) \, dx \). Simplifying gives: \( \int_{0}^{1} (3 - 3x) \, dx \). Integrate to find \( 3x - \frac{3}{2}x^2 \Big|_{0}^{1} = (3 \times 1 - \frac{3}{2} \times 1^2) - (3 \times 0 - \frac{3}{2} \times 0^2) = \frac{3}{2} \).

Key concepts

Surface Equation

The surface equation is a mathematical expression that describes a geometric shape within three-dimensional space. In this context, the given surface equation is \( f(x, y) = 2x + 2y + 2 \). This formula represents a plane. The term for each variable, \( 2x \) and \( 2y \), shows how the surface extends along the x and y axes, while the constant term 2 shifts the entire surface upward in the z-direction by two units.

In visual terms, this plane is inclined, rising steadily at angles determined by the coefficients of \( x \) and \( y \). The simplicity of this equation—being linear in both x and y—makes it easier to work with, as linear planes do not curve or bend. They extend infinitely in space, providing clear bounds for calculating areas when intersected with defined regions on the xy-plane.

Double Integral

A double integral is a powerful mathematical tool used to compute the volume under a surface over a specified region. For our exercise, we apply the double integral to find the area of the surface over the given triangular region \( R \).

We express the area as \( \int_{0}^{1} \int_{0}^{1-x} 3 \, dy \, dx \), following a two-step integration process: one with respect to y and the other with respect to x. The integrand '3' comes from evaluating the surface's derivatives, signifying a uniform stretch factor affecting the area.

The procedure involves first integrating with respect to y. Here, Y's limits are determined by the line \( y = 1-x \) within the triangle. The result of this step gives us a function of x, which we then integrate over x's range from 0 to 1, revealing the total area under our surface across the region.

Region of Integration

The region of integration is a crucial concept because it determines the limits within which we calculate the integral. In this problem, the region \( R \) is a triangle in the xy-plane with vertices at \((0,0), (1,0), \text{and } (0,1)\).

This triangle is right-angled, having its right angle at the origin \((0,0)\). The hypotenuse is described by the line \( y = 1-x \), forming the inclined boundary of the triangle. The outer boundaries lie along the x-axis and y-axis, with x ranging from 0 to 1 and y from 0 to \( 1-x \).

Recognizing the shape and boundaries of this region is essential for correctly setting the double integral's limits, ensuring that the calculation precisely covers the entire surface area resting over the triangle.

Partial Derivatives

Partial derivatives offer a way to measure how a function changes as we vary one variable while keeping the others constant. Here, we focus on the surface \( f(x, y) = 2x + 2y + 2 \).

For this exercise, we need to compute the partial derivatives \( f_x \) and \( f_y \), which are both simple to determine due to the linear nature of our surface equation:

• The partial derivative with respect to x, \( f_x \), is 2. This indicates the rate of change along the x direction.
• The partial derivative with respect to y, \( f_y \), is 2, representing the rate of change along the y direction.

We use these derivatives to adjust our integral for the surface's inclined angle. Specifically, combining them in \( \sqrt{1 + (f_x)^2 + (f_y)^2} = 3 \) modifies the area calculation to account for the surface's slope, which affects the true area seen from above.