Problem 27
Question
Integrate \(f\) over the given region. Triangle \(f(u, v)=v-\sqrt{u}\) over the triangular region cut from the first quadrant of the \(u v\) -plane by the line \(u+v=1\)
Step-by-Step Solution
Verified Answer
The integral result is \(-\frac{1}{10}\).
1Step 1: Sketch the Region of Integration
Start by identifying the given triangular region on the first quadrant. The triangle is bounded by the line \(u + v = 1\), \(u \geq 0\), and \(v \geq 0\). Plot these lines to understand the integration boundaries. The vertices of the triangle are \((0, 0)\), \((0, 1)\), and \((1, 0)\).
2Step 2: Set Up the Double Integral
The function \(f(u, v) = v - \sqrt{u}\) is to be integrated over the triangular region. Set up the double integral for \(u\) ranging from \(0\) to \(1 - v\) and \(v\) ranging from \(0\) to \(1\). The integral is given by:\[ \int_{v=0}^{1} \int_{u=0}^{1-v} (v - \sqrt{u}) \, du \, dv \]
3Step 3: Integrate with Respect to \(u\)
Compute the inner integral by integrating with respect to \(u\):\[ \int_{u=0}^{1-v} (v - \sqrt{u}) \, du \]Integrate term-by-term:\[ \int (v) \, du = vu \bigg|_{0}^{1-v} = v(1-v) \]\[ \int (-\sqrt{u}) \, du = -\frac{2}{3}u^{3/2} \bigg|_{0}^{1-v} = -\frac{2}{3}(1-v)^{3/2} \]Thus, the result of the inner integral is:\[ v(1-v) - \frac{2}{3}(1-v)^{3/2} \]
4Step 4: Integrate with Respect to \(v\)
Substitute the result from Step 3 into the outer integral:\[ \int_{v=0}^{1} \left[ v(1-v) - \frac{2}{3}(1-v)^{3/2} \right] \, dv \]This integral can be integrated term-by-term:- Integrate \(v(1-v)\): \[ \int v - v^2 \, dv = \frac{v^2}{2} - \frac{v^3}{3} \bigg|_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \]- Integrate \(-\frac{2}{3}(1-v)^{3/2}\): Use substitution method where \, let \,\( w = 1-v \), then \( dw = -dv \) \, thus \,\( \int -\frac{2}{3}w^{3/2} \, (-dw)\) \[= \left. \frac{2}{3} \cdot \frac{2}{5} (1-v)^{5/2} \right|_{0}^{1} = \frac{4}{15} \times 1 = \frac{4}{15} \]Combining both results gives:\[ \frac{1}{6} - \frac{4}{15} = \frac{5}{30} - \frac{8}{30} = -\frac{1}{10} \]
5Step 5: Conclusion
The integral calculates the area-weighted mean of the function values over the region. The computed integral value is \(-\frac{1}{10}\).
Key Concepts
Change of VariablesIntegration BoundsQuadrant RegionsIntegration Techniques
Change of Variables
When tackling a double integral, changing variables can make the integration process simpler. This technique involves substituting variables with a new set that is more convenient for the given region or problem.
For example, in some problems, using polar coordinates instead of cartesian coordinates simplifies the integration, especially if the region of integration is circular. In our exercise, however, we integrate directly in terms of \( u \) and \( v \) since the chosen variables naturally align with the triangular region's boundaries and do not require a transformation.
Nevertheless, it's always useful to sketch the region and see if a different set of variables might simplify calculations. This will often save not only time but effort too!
For example, in some problems, using polar coordinates instead of cartesian coordinates simplifies the integration, especially if the region of integration is circular. In our exercise, however, we integrate directly in terms of \( u \) and \( v \) since the chosen variables naturally align with the triangular region's boundaries and do not require a transformation.
Nevertheless, it's always useful to sketch the region and see if a different set of variables might simplify calculations. This will often save not only time but effort too!
Integration Bounds
Understanding integration bounds is crucial for setting up the double integral correctly. These bounds determine where the integration starts and ends along each axis.
In a double integral, we first fix one variable and integrate with respect to the other. Then, we change roles and integrate with respect to the first variable. For our triangular region cut by the line \( u + v = 1 \), the bounds are as follows:
In a double integral, we first fix one variable and integrate with respect to the other. Then, we change roles and integrate with respect to the first variable. For our triangular region cut by the line \( u + v = 1 \), the bounds are as follows:
- For \( u \), the integration starts from 0 and ends at \( 1 - v \).
- For \( v \), the integration ranges from 0 to 1.
Quadrant Regions
In the context of double integration, understanding quadrant regions is important when identifying the area for integration. A quadrant region refers to portions of the plane divided by the horizontal and vertical axes.
When we're operating within the first quadrant of the \( uv \)-plane, both \( u \) and \( v \) must be zero or positive. This naturally limits our discussion to the top-right portion of the plane. The given exercise restricts our exploration to this region, specifically looking at a triangular section bounded by \( u + v = 1 \).
Recognizing the active quadrant helps us accurately map the integration area, avoiding errors in the setup of integrals where negative values might lead to incorrect, misleading results.
When we're operating within the first quadrant of the \( uv \)-plane, both \( u \) and \( v \) must be zero or positive. This naturally limits our discussion to the top-right portion of the plane. The given exercise restricts our exploration to this region, specifically looking at a triangular section bounded by \( u + v = 1 \).
Recognizing the active quadrant helps us accurately map the integration area, avoiding errors in the setup of integrals where negative values might lead to incorrect, misleading results.
Integration Techniques
Choosing the right integration technique can dramatically simplify solving a double integral. In our problem, the technique involves breaking the integration process into two parts: one for \( u \) and another for \( v \).
Term-by-term integration: The method of integrating each term independently is useful when dealing with a combination of functions. In this exercise, we see the function \( v - \sqrt{u} \), where each component governs a separate part of the calculation.
Substitution: For more complex elements, substitution can help simplify the integral. Here, substitution was used for simplifying the term \( -\frac{2}{3}(1-v)^{3/2} \). This involved letting \( w = 1 - v \) and assisting with its derivative. The resulting integrals were straightforward term-wise calculations and led to deriving the solution through cumulative effort. By honing these techniques, we can solve a wide range of integration problems efficiently!
Term-by-term integration: The method of integrating each term independently is useful when dealing with a combination of functions. In this exercise, we see the function \( v - \sqrt{u} \), where each component governs a separate part of the calculation.
Substitution: For more complex elements, substitution can help simplify the integral. Here, substitution was used for simplifying the term \( -\frac{2}{3}(1-v)^{3/2} \). This involved letting \( w = 1 - v \) and assisting with its derivative. The resulting integrals were straightforward term-wise calculations and led to deriving the solution through cumulative effort. By honing these techniques, we can solve a wide range of integration problems efficiently!
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