Question

Find the area of the given surface over the region \(R\). \(f(x, y)=x^{2}+y\) over \(R,\) the triangle bounded by \(y=2 x\) \(y=0\) and \(x=2\).

Short answer

The area of the surface is \( \frac{40}{3} \).

Step-by-step solution

Understand the Boundaries of Region R

The region \( R \) is a triangle bounded by the lines \( y = 2x \), \( y = 0 \), and \( x = 2 \). This triangle lies in the \( xy \)-plane. We first determine the vertices of this triangle. The line \( y = 0 \) is the x-axis, \( y = 2x \) is a slope line, and it intersects \( x = 2 \) at the point \((2, 4)\). Thus, the vertices of the triangle are \((0, 0), (2, 0), \) and \((2, 4)\).

Set up the Double Integral

The area over which we find the surface area is in terms of \( x \) and \( y \). The expression for the surface is \( f(x, y) = x^2 + y \). We set up the double integral of \( f(x, y) \) over the region \( R \):\[\int_{0}^{2} \int_{0}^{2x} (x^2 + y) \; dy \; dx\]

Integrate with Respect to y

First, we integrate the expression \( x^2 + y \) with respect to \( y \) from 0 to \( 2x \).\[\int_{0}^{2x} (x^2 + y) \; dy = \left[ x^2y + \frac{y^2}{2} \right]_{0}^{2x} = x^2(2x) + \frac{(2x)^2}{2}\] This evaluates to:\[2x^3 + 2x^2\]

Integrate with Respect to x

Now, integrate \( 2x^3 + 2x^2 \) with respect to \( x \) from 0 to 2:\[\int_{0}^{2} (2x^3 + 2x^2) \; dx = \left[ \frac{2x^4}{4} + \frac{2x^3}{3} \right]_{0}^{2}\] This evaluates to:\[\frac{2(2)^4}{4} + \frac{2(2)^3}{3} = \frac{2 \times 16}{4} + \frac{2 \times 8}{3} = 8 + \frac{16}{3}\]\[ 8 + \frac{16}{3} = \frac{24}{3} + \frac{16}{3} = \frac{40}{3} \]

Conclude the Calculation

The final area of the surface over the triangle \( R \) is the result of the integration, which is \( \frac{40}{3} \). This area represents the accumulated value of \( f(x, y) \, \text{over} \, R \).

Key concepts

Double Integral

A double integral is a powerful mathematical tool for calculating things like areas, volumes, and other properties over a two-dimensional region. Imagine you have a surface, and you want to calculate its total height over a specific region. You use double integrals to help sum up tiny partitions of this surface within that region.
The double integral is typically written as:\[\int \int\]When finding a double integral, we usually integrate one variable at a time while treating other variables as constants. This process often involves setting up a definite integral for each variable, tailored to specific limits based on the region.

• Double integrals can be used to compute the total value of a function over an entire region.
• Doing so helps determine properties like mass, charge, and surface area.
• Understanding double integrals helps in fields like physics and engineering too.

Bounded Regions

A bounded region is an area in the plane that has boundaries or limits. When solving problems involving integrals, it's crucial to understand the boundaries of the region where you're integrating.
In our exercise, we're looking at a triangular region, which is bounded by certain lines like:

• \(y = 2x\): This line forms one boundary with a diagonal slope.
• \(y = 0\): This line is simply the x-axis, forming another boundary.
• \(x = 2\): This boundary is vertical, cutting through all other lines at the x-coordinate 2.

These boundaries are what create the contained area (or region) where the surface is being integrated. Understanding these lines will let us set the correct limits for our integration.

Triangle Vertices

Vertices are the corner points of a shape. In our triangle problem, we determined the points where two lines intersect each other, marking the vertices of our triangle. For the given region, the vertices are:

• \((0, 0)\): where \(y = 0\) and the origin meet.
• \((2, 0)\): where \(x = 2\) cuts the x-axis.
• \((2, 4)\): where \(x = 2\) intersects with the line \(y = 2x\).

Knowing these vertices is key to setting the limits in our integration. It ensures we're accurately finding areas or other properties related to our bounded region.

Integrating Functions

Integrating functions is about finding the total value a function takes over a specific interval or area. It's a way to sum up infinitely small parts to get a big picture result.
When you set up a double integral \[\int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx\]you follow these steps:

• Integrate with respect to \(y\) first. This step involves treating \(x\) as a constant while finding the antiderivative for the function related to \(y\).
• Next, integrate concerning \(x\), applying the limits determined earlier.

In our solution, we calculated first with respect to \(y\), resulting in an expression for \(x\) and then carried out the integration for \(x\).
This systematic approach helps organize and accumulate the values needed to find total properties like surface area, as shown in the step-by-step solution.