Problem 27 - University Calculus: Early Transcendentals

Question

Integrate \(f\) over the given region. Triangle \(\quad f(u, v)=v-\sqrt{u}\) over the triangular region cut from the first quadrant of the \(u v\) -plane by the line \(u+v=1\)

Step-by-Step Solution

Answer

The integral over the given region is \(-\frac{1}{10}.\)

Step 1: Visualize the Region

The problem involves a triangular region in the first quadrant, bounded by the line \(u + v = 1\). This line intersects the u-axis at \((1,0)\) and the v-axis at \((0,1)\). The vertices of the triangular region are \((0,1)\), \((1,0)\), and \((0,0)\).

Step 2: Set Up the Double Integral

We need to integrate the function \(f(u, v) = v - \sqrt{u}\) over the triangular region. By expressing the bounds for \(v\) as dependent on \(u\), we see that for a given \(u\), \(v\) ranges from 0 to \(1-u\). Thus, the double integral is set up as: \[ \int_{0}^{1} \int_{0}^{1-u} (v - \sqrt{u}) \, dv \, du. \]

Step 3: Integrate with respect to v

Integrate the inner integral with respect to \(v\): \(\int_{0}^{1-u} (v - \sqrt{u}) \, dv\). This computes to: \[ \left[ \frac{v^2}{2} - v\sqrt{u} \right]_{0}^{1-u} = \frac{(1-u)^2}{2} - (1-u)\sqrt{u}. \]

Step 4: Simplify and Integrate with respect to u

Substitute the result from Step 3 into the outer integral and simplify: \[ \int_{0}^{1} \left( \frac{(1-u)^2}{2} - (1-u)\sqrt{u} \right) \, du. \] Simplify and separate the integral: \[ \frac{1}{2} \int_{0}^{1} (1 - 2u + u^2) \, du - \int_{0}^{1} (1-u)\sqrt{u} \, du. \]

Step 5: Solve the Integrals

1. Solve \(\frac{1}{2} \int_{0}^{1} (1 - 2u + u^2) \, du\) which results in \(\frac{1}{2} \left[ u - u^2 + \frac{u^3}{3} \right]_0^1 = \frac{1}{2} \left( 1 - 1 + \frac{1}{3} \right) = \frac{1}{6}.\) 2. Solve \(\int_{0}^{1} (1-u)\sqrt{u} \, du\): let \(u = t^2\), then \(du = 2t \, dt\). The boundaries update to 0 to 1: \[ 2 \int_{0}^{1} (1-t^2) t^2 \, dt = 2 \int_{0}^{1} (t^2 - t^4) \, dt. \] This resolves to \[ 2 \left[ \frac{t^3}{3} - \frac{t^5}{5} \right]_0^1 = 2 \left( \frac{1}{3} - \frac{1}{5} \right) = 2 \times \frac{2}{15} = \frac{4}{15}. \]

Step 6: Compute the Final Result

Combine the results from the previous step: \(\frac{1}{6} - \frac{4}{15} = \frac{5}{30} - \frac{8}{30} = -\frac{1}{10}.\) Hence, the integral of \(f(u, v) = v - \sqrt{u}\) over the given region is \(-\frac{1}{10}.\)

Key Concepts

Triangular Region

Understanding the triangular region in integration problems is crucial as it defines the limits for integration. In this exercise, the triangular region is situated in the first quadrant of the uv-plane. The triangle is formed by the line equation \(u + v = 1\) which crosses the axes, creating a boundary for our problem. The vertices of this region are \((0,0)\), \((1,0)\), and \((0,1)\), forming a right-angled triangle. Knowing these vertices helps in setting up the limits for our integrals. Visualizing this triangle on the coordinate plane can aid in comprehending the region you are working within, enabling you to better set the integration bounds.

First Quadrant

The first quadrant is an essential part of Cartesian coordinates where both the \(u\) and \(v\) values are positive. This positive nature constraints our function within this boundary and ensures that our variables stay positive during integration. Within this first quadrant, the line \(u + v = 1\) helps in establishing our region of integration, essentially slicing a piece of the first quadrant as our area of interest. Utilizing constraints like those provided by the first quadrant can simplify the problem as it can prevent negative values which are generally not practical in many real-world problems.

Changing Variables

Changing variables, particularly in integration, is a helpful technique in simplifying complex integrals. During the integration in this exercise, a substitution is used for the term \((1-u)\sqrt{u}\). Letting \(u = t^2\), and thereby \(du = 2t\, dt\), allows us to convert the integral into a more straightforward form. This new form can be easier to solve, especially when dealing with polynomials or radicals. Substitution helps break down difficult integrals into manageable parts by transforming them into complete squares or bringing them into a form where standard integrals can be applied.

Integral Computation

Integral computation involves taking the function you want to integrate and finding the area under it within specified boundaries. In this exercise, the function \(f(u,v) = v - \sqrt{u}\) is integrated over a triangular region. The process requires setting up a double integral where the ranges are dictated by our triangle, first integrating with respect to \(v\) and then \(u\). The integration steps make use of both definite and indefinite integral calculations to sum the function's area over its region. Simplifying inside integrals first before moving on to the outer integral is important for efficient calculation. Finally, you gather your computed values such as \(\frac{1}{6}\) and subtract \(\frac{4}{15}\), ending with the result of \(-\frac{1}{10}\). This sequence showcases the importance of performing each step carefully and methodically to ensure accuracy throughout the process.