Problem 38 - Calculus Early Transcendentals

Question

Sketch each region and write an iterated integral of a continuous function \(f\) over the region. Use the order \(d x d y\). The region in the first quadrant bounded by the \(x\) -axis, the line \(x=6-y,\) and the curve \(y=\sqrt{x}\)

Step-by-Step Solution

Answer

Question: Set up an iterated integral for the continuous function, \(f\), over the region in the first quadrant bounded by the x-axis, the line \(x=6-y\), and the curve \(y=\sqrt{x}\). Use the order \(dx\, dy\). Answer: $$\int_{0}^{6} \int_{0}^{\sqrt{x}} f(x, y) \, dy\, dx$$

Step 1: Sketch the Region

Let's start by sketching the given region. First, draw the x-axis and the line \(x=6-y\). Then, sketch the curve \(y=\sqrt{x}\). You will notice that the region is in the first quadrant and is enclosed by the three mentioned boundaries.

Step 2: Determine Intersection Points

Next, we need to determine the intersection points of our given boundaries. 1. Intersection point between line \(x=6-y\) and curve \(y=\sqrt{x}\): \((6-y)^2 = y \Rightarrow y^2-12y+36=y \Rightarrow y^2-13y+36=0\). The solutions to this quadratic equation are \(y=4\) and \(y=9\). For \(y=4\), \(x=(6-4)=2\), and for \(y=9\), \(x=(6-9)=-3\). Since we are in the first quadrant, only the point \((2,4)\) is relevant. 2. Intersection point of curve \(y=\sqrt{x}\) and x-axis: \(y=0 \Rightarrow x=0^2=0\). 3. Intersection point of line \(x=6-y\) and x-axis: \(y=0 \Rightarrow x=6-0=6\). So, the intersection points are \((0,0)\), \((2,4)\), and \((6,0)\).

Step 3: Determine Integral Limits

Based on the intersection points, we can now determine the limits for our iterated integral for function \(f\). For \(y\) limits: \(0 \leq y \leq \sqrt{x}\) (x-axis to the curve \(y=\sqrt{x}\)) For \(x\) limits: \(6-y \leq x \leq 6\) (bounded by the line \(x=6-y\) and \(x=6\))

Step 4: Write the Iterated Integral

Now, we can write the iterated integral for the continuous function \(f\): $$\int_{0}^{6} \int_{0}^{\sqrt{x}} f(x, y) \, dy\, dx$$

Key Concepts

Continuous Function

A continuous function is a mathematical concept that ensures there is no interruption in the value of the function as we move along its graph. In the context of iterated integrals, working with a continuous function over a pointed region makes the integration process smooth and unambiguous since the function has no sudden jumps or holes.

During the integration of a continuous function, predicted behaviors, such as gradual increases or decreases in rates, allow for the precise calculation of the area under the curve within a specific region. Continuous functions typically represent real-world phenomena where quantities change smoothly over time or space, providing a realistic model for many practical problems.

Region Sketching

Region sketching involves drawing the area on a graph where an integral will be evaluated. It is an essential visual aid that helps identify the boundaries and shape of the region involved in a problem.

For an iterated integral, sketching the region can reveal the appropriate limits of integration. By plotting the functions that bound the region, such as the x-axis, the line x=6-y, and the curve y=\(\backslash sqrt{x}\) for the given exercise, we can observe how these boundaries intersect. In the exercise, this step allowed us to see a bounded area in the first quadrant, enclosed by the mentioned lines and curve, crucial for setting up the integration process correctly.

Integration Limits

Setting up integration limits is a critical step in solving an iterated integral, as it defines the range over which the function is to be integrated. These limits are derived from the region's boundaries.

For the exercise in question, the y limits range from the x-axis (y=0) to the curve y=\(\backslash sqrt{x}\), while the x limits run from the vertical line x=6-y to the vertical line x=6 on a rightward trajectory. Choosing accurate integration limits ensures that the area calculated covers the entire region of interest and no more. Moreover, recognizing which variable to integrate with respect to first is guided by the order of integration specified, in this case, dxdy.

Quadratic Equation Solutions

Quadratic equations appear frequently in integration problems when finding the points of intersection between curves. These solutions are necessary for determining the limits of integration.

In our exercise, encountering y^2-13y+36=0 led to the need for finding the quadratic solutions which are y=4 and y=9. However, since we are focused on the first quadrant, we discard the solution leading to a negative x-coordinate, keeping only the intersection point at (2,4). These solutions are crucial, as incorrectly calculated or interpreted intersection points could lead to incorrect integration limits, thus affecting the final result for the iterated integral.