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Chapter 15: Problem 7

Evaluate the double integral. $$\iint_{D} y^{2} d A, \quad D=\\{(x, y) |-1 \leqslant y \leqslant 1,-y-2 \leqslant x \leqslant y\\}$$

Short Answer

Expert verified
The double integral evaluates to \( \frac{4}{3} \).

Step by step solution

01

Determine the Integration Limits

First, we look at the region \( D \) defined by \(-1 \leq y \leq 1\) and \(-y-2 \leq x \leq y\). This gives us our limits for \( y \) and consequently for \( x \) for any fixed \( y \). For each fixed \( y \), \( x \) ranges from \(-y - 2\) to \( y \).
02

Set Up the Double Integral

Based on the integration limits from Step 1, we set up the double integral as follows: \[ \int_{-1}^{1} \int_{-y-2}^{y} y^2 \, dx \, dy \]. This formulation maintains \( y^2 \) as the integrand, which is independent of \( x \).
03

Integrate with respect to x

Integrate the inner integral \( \int_{-y-2}^{y} y^2 \, dx \), treating \( y^2 \) as a constant with respect to \( x \). The result is \( y^2 \cdot (x) \bigg|_{-y-2}^{y} \). Evaluating this, we get \( y^2 (y - (-y-2)) = y^2 (2y + 2) = y^2 (2(y+1)) = 2y^2(y+1) \).
04

Integrate with respect to y

Now, integrate the expression \( 2y^2(y+1) \) with respect to \( y \) over \([-1, 1]\): \[ \int_{-1}^{1} 2y^2(y+1) \, dy = 2 \int_{-1}^{1} (y^3 + y^2) \, dy \]. Find each integral separately: 1. \( \int_{-1}^{1} y^3 \, dy = 0 \) due to symmetry of an odd function over symmetric limits.2. \( \int_{-1}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-1}^{1} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}\).
05

Combine and Simplify

Combine the results of the integrals from Step 4: \( 2 \left( 0 + \frac{2}{3} \right) = \frac{4}{3} \). Thus, evaluating the double integral results in \( \frac{4}{3} \).

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Key Concepts

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

Integration Limits
Understanding integration limits is a fundamental aspect of solving double integrals. In this exercise, the region \( D \) is defined by the inequalities \(-1 \leq y \leq 1\) and \(-y-2 \leq x \leq y\). These provide the boundaries or limits across which the integration takes place.
  • The first set of limits, \(-1 \leq y \leq 1\), defines the range for the outer integral, representing the vertical span of the region \( D \).
  • The second set, \(-y-2 \leq x \leq y\), specifies the horizontal extent for each fixed value of \( y \). This range serves as the limits for the inner integral.
By clearly identifying these limits, you ensure the double integral correctly represents the area under consideration.
Inner Integral
The inner integral is the first integration that occurs in a double integral setup, focusing on one variable while regarding others as constants. In our example, we solve \( \int_{-y-2}^{y} y^2 \, dx \).

  • The function \( y^2 \) is independent of \( x \), which simplifies the evaluation of this integral. Hence, \( y^2 \) can be treated as a constant with respect to \( x \).
  • Integration of a constant with respect to \( x \) simply involves multiplying by the range of \( x \), i.e., \( (x) \bigg|_{-y-2}^{y} \).
Evaluating gives \( y^2 \cdot (2(y+1)) = 2y^2(y+1) \), setting the stage for the outer integration.
Outer Integral
Once the inner integral is solved, we progress to the outer integral, which handles the remaining variable. Given the result from the inner integral, \( 2y^2(y+1) \), the new integral is \( \int_{-1}^{1} 2y^2(y+1) \, dy \).

This step simplifies the problem to integrating a polynomial, breaking down further as:
  • \( \int_{-1}^{1} 2(y^3 + y^2) \, dy \), which can be split into two separate integrals for more straightforward evaluation.
Successfully solving these integrals brings us closer to evaluating the full double integral.
Symmetric Integration
A key consideration in this problem is recognizing symmetry in the integrand. Sometimes, symmetry can simplify calculations significantly.

For instance, the term \( y^3 \) in the polynomial \( y^3 + y^2 \) is an odd function. Integrating an odd function over symmetric limits (from \(-1\) to \(1\)) results in zero:
  • \( \int_{-1}^{1} y^3 \, dy = 0 \)
Leveraging symmetry reduces the complexity of the integral, allowing us to focus on more significant terms, like \( y^2 \), that contribute to the final result.
Integral Evaluation
The culmination of a double integral involves combining everything to reach an evaluation. After determining the symmetric properties and evaluating each term separately, the integrals can be summed.

  • After evaluating \( \int_{-1}^{1} y^3 \, dy = 0 \) and \( \int_{-1}^{1} y^2 \, dy = \frac{2}{3} \), we multiply by the factor from the setup: \( 2 \).
  • Thus, the combined result of the double integration becomes \( 2 \left( 0 + \frac{2}{3} \right) = \frac{4}{3} \).
This final value represents the area under the function \( y^2 \) over the defined region \( D \). Understanding these steps ensures clarity in the double integration process.

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Most popular questions from this chapter

Evaluate the integral by making an appropriate change of variables. \(\iint_{R}(x+y) e^{x^{2}-y^{2}} d A,\) where \(R\) is the rectangle enclosed by the lines \(x-y=0, x-y=2, x+y=0,\) and \(x+y=3\)

\(21-34\) Use spherical coordinates. Evaluate \(\iiint_{E} e^{\sqrt{x^{2}+y^{2}+z^{2}}} d V,\) where \(E\) is enclosed by the sphere \(x^{2}+y^{2}+z^{2}=9\) in the first octant.

Set up, but do not evaluate, integral expressions for (a) the mass, (b) the center of mass, and (c) the moment of inertia about the \(z\) -axis. The hemisphere \(x^{2}+y^{2}+z^{2} \leqslant 1, z \geqslant 0\) \(\rho(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}\)

Evaluate the integral by reversing the order of integration. $$\int_{0}^{1} \int_{x}^{1} e^{x / y} d y d x$$

\(\begin{array}{l}{\text { (a) Use cylindrical coordinates to show that the volume of }} \\ {\text { the solid bounded above by the sphere } r^{2}+z^{2}=a^{2} \text { and }} \\ {\text { below by the cone } z=r \cot \phi_{0}\left(\text { or } \phi=\phi_{0}\right), \text { where }} \\\ {0<\phi_{0}<\pi / 2, \text { is }}\end{array}\) $$V=\frac{2 \pi a^{3}}{3}\left(1-\cos \phi_{0}\right)$$ (b) Deduce that the volume of the spherical wedge given by \(\rho_{1} \leqslant \rho \leqslant \rho_{2}, \theta_{1} \leqslant \theta \leqslant \theta_{2}, \phi_{1} \leqslant \phi \leqslant \phi_{2}\) is $$\Delta V=\frac{\rho_{2}^{3}-\rho_{1}^{3}}{3}\left(\cos \phi_{1}-\cos \phi_{2}\right)\left(\theta_{2}-\theta_{1}\right)$$ (c) Use the Mean Value Theorem to show that the volume in part (b) can be written as $$\Delta V=\tilde{\rho}^{2} \sin \tilde{\phi} \Delta \rho \Delta \theta \Delta \phi$$ where \(\tilde{\rho}\) lies between \(\rho_{1}\) and \(\rho_{2}, \tilde{\phi}\) lies between \(\phi_{1}\) and \(\phi_{2}, \Delta \rho=\rho_{2}-\rho_{1}, \Delta \theta=\theta_{2}-\theta_{1},\) and \(\Delta \phi=\phi_{2}-\phi_{1}\)
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