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Chapter 5: Problem 77

In the following exercises, evaluate the double integral \(\iint_{D} f(x, y) d A\) over the region \(D .\) \(f(x, y)=x y\) and \(D=\left\\{(x, y) |-1 \leq y \leq 1, y^{2}-1 \leq x \leq \sqrt{1-y^{2}}\right\\}\)

Short Answer

Expert verified
The double integral evaluates to 0.

Step by step solution

01

Identify the Limits of Integration

The region \(D\) is bounded as \(-1 \leq y \leq 1\) and \(y^2 - 1 \leq x \leq \sqrt{1 - y^2}\). Since the limits for \(y\) are constants and for \(x\) are functions of \(y\), we will integrate with respect to \(x\) first and then \(y\).
02

Set Up the Double Integral

Write the double integral as follows:\[\int_{-1}^{1} \int_{y^2 - 1}^{\sqrt{1 - y^2}} xy \, dx \, dy\]The outer integral runs from -1 to 1 with respect to \(y\), and the inner integral runs from \(y^2 - 1\) to \(\sqrt{1 - y^2}\) with respect to \(x\).
03

Integrate with Respect to \(x\)

Integrate \(xy\) with respect to \(x\):\[\int (xy) \, dx = \frac{x^2}{2}y\]Evaluate this from \(x = y^2 - 1\) to \(x = \sqrt{1 - y^2}\):\[\frac{(\sqrt{1 - y^2})^2}{2}y - \frac{(y^2 - 1)^2}{2}y\]Simplify this to find:\[\int_{y^2 - 1}^{\sqrt{1 - y^2}} xy \, dx = \frac{1 - y^2 - (y^2 - 1)^2}{2} y\]
04

Simplify Inner Integration Result

Simplify the expression obtained after integrating with respect to \(x\):\[\frac{1 - y^2 - (y^4 - 2y^2 + 1)}{2} y = \frac{2 - 2y^2 - y^4}{2} y\]This simplifies further to:\[\frac{(2 - y^4 - 2y^2)}{2} y\]
05

Integrate with Respect to \(y\)

Now integrate the expression over the interval \(-1 \leq y \leq 1\):\[\int_{-1}^{1} \frac{(2 - y^4 - 2y^2)}{2} y \, dy\]Distribute \(y\) across the polynomial and break into separate integrals:\[\frac{1}{2} \int_{-1}^{1} (2y - 2y^3 - y^5) \, dy\]Integrate term-by-term:\[\frac{1}{2} \left[ y^2 - \frac{1}{2} y^4 - \frac{1}{6} y^6 \right]_{-1}^{1}\]
06

Evaluate the Definite Integral

Evaluate the antiderivative from \(y = -1\) to \(y = 1\):\[\frac{1}{2} \left( [1 - \frac{1}{2} \cdot 1^2 - \frac{1}{6} \cdot 1^6] - [-1 + \frac{1}{2} \cdot (-1)^2 + \frac{1}{6} \cdot (-1)^6] \right)\]Simplifying gives:\[\frac{1}{2} (\frac{6}{6} - \frac{3}{6} - \frac{1}{6}) - (-\frac{6}{6} + \frac{3}{6} + \frac{1}{6})\]This evaluates to:\(0\)
07

Conclusion

The value of the double integral \(\iint_{D} f(x, y) \, dA\) over the specified region \(D\) is 0. This confirms the symmetry of the function and region about the \(x\)-axis, leading to the cancellation of positive and negative areas in the computation.

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Key Concepts

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

Limits of Integration
In a double integral, the limits of integration play a crucial role. They define the boundaries within which the integration takes place. For the given exercise, we have two sets of limits: for the variable \(y\), the limits are constant, and for the variable \(x\), the limits are functions of \(y\). This tells us the order in which we should integrate. Here, it's allowed us to integrate with respect to \(x\) first, then \(y\).

Why the Order of Integration Matters
- The order of integration is determined by how the limits are expressed.
- If functions are nested within one another, our integration method will often begin with the innermost nested variable.
- Correct order ensures we do not exceed the region described by the limits.

Here, we notice that \(x\) has limits \(y^2 - 1\) to \(\sqrt{1 - y^2}\). Meanwhile, \(y\) spans from \(-1\) to \(1\). This order effectively "layers" the region of integration, examining vertical slices across the \(x\) range defined by \(y\).

Steps to Set Up Limits
1. Identify constant limits.
2. Determine which variable is dependent. Set corresponding limits.
3. Confirm that the order of integration respects the nested functions within the limits.
Region of Integration
The region of integration describes the specific area over which the double integral is evaluated. It is defined both graphically and algebraically. In this exercise, the region \(D\) is described by the inequalities for \(x\) and \(y\). It's essential to visualize this region since it confirms how the function behaves over specified intervals.

Understanding the Region with Inequalities
- The inequality \(-1 \leq y \leq 1\) indicates the strip in the vertical direction, spanning from \(y = -1\) to \(y = 1\).
- For \(x\), \(y^2 - 1 \leq x \leq \sqrt{1-y^2}\) defines arcs or curves on the \(xy\)-plane. This effectively "caps" the values of \(x\) according to the value of \(y\).

Visualizing the Region
- Sketching the region by first plotting \(y = -1\) and \(y = 1\) as horizontal lines helps.
- Understanding \(x = y^2 - 1\) and \(x = \sqrt{1-y^2}\) shapes can highlight the boundary curves 鈥� a parabola opening to the left and a semicircle opening to the right, respectively.
- The overlap of these graphs between \(y = -1\) and \(y = 1\) defines the actual region \(D\).

Visually, this region \(D\)'s characteristics might suggest symmetry which can further aid in easier computation or fact-checking our integral results.
Symmetry in Integrals
Recognizing symmetry can dramatically simplify the evaluation of integrals. Symmetry involves the cancellation of areas above and below a given axis. In this problem, evaluating the double integral over the specified region \(D\) gives a result of zero, indicating symmetry.

How Symmetry Simplifies Integrals
- Detecting symmetry can tell us that parts of our region "cancel out" regarding the function we integrate over.
- If a function or region is symmetrical around an axis, such as the \(x\)-axis, portions on either side may equalize to zero when integrated.

Identifying Symmetry in This Problem
- The function \(f(x, y) = xy\) is odd with respect to both variables. This means flipping signs of either variable also flips the sign of the function. This trait can lead to symmetry in results.
- Since \(D\) is balanced around the \(x\)-axis due to its bounds, this symmetry cancels out the integral.

Practical Use of Symmetry
- Simplifies integration by avoiding direct computation of complex regions.
- Allows for quicker guesses and checks, especially when symmetry is glaringly apparent in inequalities.
- Reduces risk of calculation errors due to overlooked symmetry properties during initial calculations.

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Most popular questions from this chapter

\(f(x, y, z)=\sqrt{x^{2}+y^{2}}, \quad B\) is bounded above by the half-sphere \(x^{2}+y^{2}+z^{2}=9\) with \(z \geq 0\) and below by the cone \(2 z^{2}=x^{2}+y^{2}\)

[T] The average density of a solid \(Q\) is defined as \(\rho_{\text {ave}}=\frac{1}{V(Q)} \iiint_{Q} \rho(x, y, z) d V=\frac{m}{V(Q)}, \quad\) where \(\quad V(Q)\) and \(m\) are the volume and the mass of \(Q,\) respectively. If the density of the unit ball centered at the origin is \(\rho(x, y, z)=e^{-x^{2}-y^{2}-z^{2}},\) use a CAS to find its average density. Round your answer to three decimal places.

The following problems examine Mount Holly in the state of Michigan. Mount Holly is a landfill that was converted into a ski resort. The shape of Mount Holly can be approximated by a right circular cone of height 1100 ft and radius 6000 \(\mathrm{ft.}\) If the compacted trash used to build Mount Holly on average has a density 400 \(\mathrm{lb} / \mathrm{ft}^{3}\) , find the amount of work required to build the mountain.

The solid \(Q=\left\\{(x, y, z) | 0 \leq x^{2}+y^{2} \leq 16, x \geq 0, y \geq 0,0 \leq z \leq x\right\\}\) has the density \(\rho(x, y, z)=k .\) Show that the moment \(M_{x y}\) about the \(x y\) -plane is half of the moment \(M_{y z}\) about the \(y z\) -plane.

\([\mathbf{T}]\) a. Evaluate the integral \(\iiint_{E} e^{\sqrt{x^{2}+y^{2}+z^{2}}} d V\) where \(E\) is bounded by the spheres \(4 x^{2}+4 y^{2}+4 z^{2}=1\) and \(x^{2}+y^{2}+z^{2}=1\) b. Use a CAS to find an approximation of the previous integral. Round your answer to two decimal places.
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