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Chapter 15: Problem 27

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 \(u+v=1\)

Short Answer

Expert verified
The integral result is \(-\frac{1}{10}\).

Step by step solution

01

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 \((1, 0)\).
02

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:\[ \int_{v=0}^{1} \int_{u=0}^{1-v} (v - \sqrt{u}) \, du \, dv \]
03

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:\[ v(1-v) - \frac{2}{3}(1-v)^{3/2} \]
04

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:\[ \frac{1}{6} - \frac{4}{15} = \frac{5}{30} - \frac{8}{30} = -\frac{1}{10} \]
05

Conclusion

The integral calculates the area-weighted mean of the function values over the region. The computed integral value is \(-\frac{1}{10}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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!
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:
  • For \( u \), the integration starts from 0 and ends at \( 1 - v \).
  • For \( v \), the integration ranges from 0 to 1.
These bounds reflect the triangular area within the first quadrant, where both variables are always non-negative. Carefully considering the bounds ensures that the entire area of the triangle is covered during integration.
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.
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!

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Most popular questions from this chapter

The Parallel Axis Theorem Let \(L_{\mathrm{cm}}\) be a line through the center of mass of a body of mass \(m\) and let \(L\) be a parallel line \(h\) units away from \(L_{c . m .}\) The Parallel Axis Theorem says the moments of inertia \(I_{\mathrm{cm}}\) and \(I_{L}\) of the body about \(L_{\mathrm{cm}}\) and \(L\) satisfy the equation $$ I_{L}=I_{\mathrm{c.m.}}+m h^{2} $$ As in the two-dimensional case, the theorem gives a quick way to calculate one moment when the other moment and the mass are known. Proof of the Parallel Axis Theorem a. Show that the first moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the \(y z\) -plane. What does the formula \(\overline{x}=M_{y z} / M\) then tell you?) b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line \(L_{c . m . \text { along the }}\) \(z\) z-axis and the line \(L\) perpendicular to the \(x y\) -plane at the point \((h, 0,0) .\) Let \(D\) be the region of space occupied by the body. Then, in the notation of the figure, $$ I_{L}=\iiint_{D}|\mathbf{v}-h \mathbf{i}|^{2} d m $$ Expand the integrand in this integral and complete the proof.

In Exercises \(33-46,\) sketch the region of integration and write an equivalent double integral with the order of integration reversed. \begin{equation} \int_{0}^{1} \int_{2}^{4-2 x} d y d x \end{equation}

Average distance to a given point inside a disk Let \(P_{0}\) be a point inside a circle of radius \(a\) and let \(h\) denote the distance from \(P_{0}\) to the center of the circle. Let \(d\) denote the distance from an arbitrary point \(P\) to \(P_{0} .\) Find the average value of \(d^{2}\) over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and \(P_{0}\) on the \(x\) -axis.)

In Exercises \(49-52,\) use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. $$ \begin{array}{l}{\text { a. Plot the Cartesian region of integration in the } x y \text { -plane. }} \\ {\text { b. Change each boundary curve of the Cartesian region in part }} \\ {\text { (a) to its polar representation by solving its Cartesian equation for } r \text { and } \theta .}\end{array} $$ $$ \begin{array}{l}{\text { c. Using the results in part (b), plot the polar region of integra- }} \\ {\text { tion in the } r \theta \text { -plane. }} \\\ {\text { d. Change the integrand from Cartesian to polar coordinates. De- }} \\ {\text { termine the limits of integration from your coordinates. De- }} \\ {\text { evaluate the polar integral using the CAS integration utility. }}\end{array} $$ $$ \int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} d y d x $$

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \begin{equation} \int_{0}^{2} \int_{0}^{4-x^{2}} \frac{x e^{2 y}}{4-y} d y d x \end{equation}
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