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Chapter 7: Problem 32

Evaluate the given double integral for the specified region \(R\).\(\iint_{R} 2 x d A\), where \(R\) is the region bounded by \(y=\frac{1}{x^{2}}, y=x\), and \(x=2\).

Short Answer

Expert verified
\( \frac{14}{3} - 2 \ln 2 \)

Step by step solution

01

- Understand the region R

The region R is bounded by the curves: \( y = \frac{1}{x^2} \), \( y = x \), and \( x = 2 \). We first need to determine the points of intersection.
02

- Find points of intersection

Set \( y = \frac{1}{x^2} \) equal to \( y = x \) to find their intersections: \( x = \frac{1}{x^2} \). Thus, \( x^3 = 1 \) so \( x = 1 \). Therefore, the points of intersection are: \( (1,1) \) and \( (2,2) \).
03

- Setup the integral

Since \( R \) is vertically simple (bounded between the curves in y and limits in x), the double integral can be expressed as: \[ \int_{1}^{2} \int_{\frac{1}{x^2}}^{x} 2x \, dy \, dx \]
04

- Evaluate the inner integral

First, integrate with respect to \( y \): \[ \int_{\frac{1}{x^2}}^{x} 2x \, dy = 2x \cdot ( y \big|_{\frac{1}{x^2}}^{x} ) = 2x \left( x - \frac{1}{x^2} \right) = 2x \left( x - x^{-2} \right) = 2x^2 - 2x^{-1} \]
05

- Integrate with respect to x

Now, evaluate the integral with respect to \( x \): \[ \int_{1}^{2} (2x^2 - 2x^{-1}) \, dx \] Compute the integrals separately: \[ \int 2x^2 \, dx = \frac{2}{3}x^3 \] and \[ \int -2x^{-1} \, dx = -2 \ln|x| \]Thus, \[ \left. \frac{2}{3}x^3 - 2 \ln|x| \right|_{1}^{2} = \left( \frac{2}{3}(2)^3 - 2 \ln|2| \right) - \left( \frac{2}{3}(1)^3 - 2 \ln|1| \right) \]
06

- Compute the definite integral

Substitute the bounds: \[ \left( \frac{2}{3} \times 8 - 2 \ln 2 \right) - \left( \frac{2}{3} - 0 \right) = \left( \frac{16}{3} - 2 \ln 2 \right) - \frac{2}{3} = \frac{16}{3} - \frac{2}{3} - 2 \ln 2 = \frac{14}{3} - 2 \ln 2 \]

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Key Concepts

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

integration bounds
When evaluating a double integral, the integration bounds are crucial. They determine the region over which the integral is calculated. In this exercise, the region R is bounded by the curves: \( y = \frac{1}{x^2} \), \( y = x \), and \( x = 2 \).
intersection points
Intersection points between curves help us define the boundaries for our integral. To find where \( y = \frac{1}{x^2} \) and \( y = x \) intersect, we set these equations equal: \( x = \frac{1}{x^2} \). Solving for \( x \), we get \( x^3 = 1 \), so \( x = 1 \). Hence, the points of intersection are \( (1,1) \) and \( (2,2) \).
vertical strip method
The vertical strip method involves slicing our region into narrow vertical strips. For each \( x \) value from \( 1 \) to \( 2 \), we integrate \( y \) from \( \frac{1}{x^2} \) to \( x \). This translates into setting up our double integral as: \[ \int_{1}^{2} \int_{\frac{1}{x^2}}^{x} 2x \, dy \, dx \]
inner integral
The inner integral is evaluated first and deals with the variable \( y \). Here, we integrate \( 2x \) with respect to \( y \), with bounds \( y = \frac{1}{x^2} \) to \( y = x \). This means: \int_{\frac{1}{x^2}}^{x} 2x \, dy = 2x \left( y \Big|_{\frac{1}{x^2}}^{x} \right) = 2x \left( x - \frac{1}{x^{2}}\right) = 2x(x - x^{-2}) = 2x^2 - 2x^{-1}.
logarithmic integration
After solving the inner integral, we integrate with respect to \( x \). This involves computing two separate integrals: \int_{1}^{2} 2x^2 \, dx and \int_{1}^{2} -2x^{-1} \, dx. The first integral is straightforward: \int 2x^2 \, dx = \frac{2}{3}x^3. For the second one, recall that \int x^{-1} = \ln|x|, giving us: \int -2x^{-1} \, dx = -2 \ln|x|. Combining both, we get the final expression: \left. \frac{2}{3}x^3 - 2 \ln|x| \right|_{1}^{2}.

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Most popular questions from this chapter

Use inequalities to describe \(R\) in terms of its vertical and horizontal cross sections.\(R\) is the triangle with vertices \((1,0),(1,1)\), and \((2,0)\).

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