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For Problems \(39-42,\) sketch the region of integration then rewrite the integral with the order of integration reversed. $$\int_{0}^{2} \int_{r-2}^{2-y} f(x, y) d x d y$$

Short Answer

Expert verified
Reverse the order to integrate from \(x=r-2\) to \(2\), then \(y=x+r-2\) to \(2-x\).

Step by step solution

01

Understand the given integral limits

The problem involves a double integral: \(\int_{0}^{2} \int_{r-2}^{2-y} f(x, y) \, dx \, dy\). The limits for \(x\) are from \(x = r-2\) to \(x = 2-y\), and \(y\) ranges from \(y = 0\) to \(y = 2\). These limits define a region in the xy-plane.
02

Sketch the region of integration

To sketch the region, note the conditions given by the limits:- The curve \(x = r-2\) is constant if \(r\) is a given value or parameter. If unspecified, interpret as \(x = -2\).- \(x = 2-y\) describes a line. Solve for \(y\): \(y = 2 - x\).Within the bounds \(0 \leq y \leq 2\), \(x\) varies from \(r-2\) to \(2-y\). Sketch lines \(y = 0\), \(y = 2\), \(x = r-2\), and \(y = 2-x\). The region is the area of overlap between these curves and lines.
03

Determine the new order of integration

The current order of integration is \(dx \, dy\). To reverse, consider the alternative limits:- Identify the range for \(x\): From the minimum \(x = r-2\) to maximum \(x = 2\).- Solve \(2-y = x\): \(y = 2 - x\).- For a given \(x\), \(y\) ranges between \(y = x+r-2\) (from the initial \(x=r-2\) condition) and \(y = 2-x\). Thus, for \(x\)'s in these limits, \(y\) ingrowns from \(x+r-2\) to \(2-x\).
04

Update the integral with reversed limits

With the new integration order established as \(dy \, dx\), rewrite the integral:\[\int_{r-2}^{2} \int_{x+r-2}^{2-x} f(x, y) \, dy \, dx\]This integral respects the previously sketched region of integration, but reverses the order to integrate \(y\) first, then \(x\).

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Key Concepts

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

Double Integrals
Double integrals are a part of multivariable calculus, used to compute the volume under a surface in a given region in the xy-plane. When dealing with functions of two variables, as in our exercise, the double integral helps us calculate the accumulation of quantities over a specified area. This is particularly invaluable in physics and engineering to find quantities such as mass, center of mass, or electric charge over a region.

The notation \( \int \int f(x, y) \, dx \, dy \) represents two integrations: one over the \(x\) variable and another over the \(y\) variable. The order of these integrations can be interchanged under proper conditions, which involves reversing the order of integration limits. This typically requires a careful analysis of the geometric region of integration. Changing the order can simplify the computation significantly depending on the function or region involved.

When you perform a double integral, you are effectively slicing through the region of interest and accumulating the value computed by the function \(f(x, y)\) over each small slice. In our specific problem, reversing the order can provide a new perspective on the region, often leading to an easier or more intuitive integral to solve.
Region of Integration
In the context of double integrals, the region of integration is the area over which you are evaluating your integral. By understanding this region geometrically, you can accurately set up your limits for integration. In the exercise we're examining, the given integral [\( \int_{0}^{2} \int_{r-2}^{2-y} f(x, y) \, dx \, dy \)] initially defines the region by the limits of \(x\) and \(y\).

To sketch and identify the region, look at the boundary conditions:
  • \(y = 0\) and \(y = 2\) define the horizontal bounds of the region.
  • \(x = r-2\) is a vertical line, while \(x = 2-y\) represents a diagonal line.
The intersection of these lines and curves determines the enclosed area, which is the region of integration. By understanding the geometric configuration, you can analyze how reversing the limits affects the integral.

Visualizing this involves drawing the lines on the xy-plane, finding points of intersection, and shading the overlapping areas. This shaded area is where the integration will occur, and recognizing it allows one to properly switch the integration limits when needed.
Integration Limits
Integration limits define the bounds over which variables extend during integration. These limits are crucial because they set the boundaries for computing a double integral. Each variable in a double integral has its own set of integration limits, which describe the extent of the region in one direction.

In the original problem, the initial integration order of \(dx \, dy\) was altered by reversing these limits. This involved:
  • Identifying the new range for \(x\), from \(x = r-2\) to \(x = 2\), based on the earlier integration bounds change.
  • Re-evaluating the limits of \(y\) for a given \(x\), ranging between \(y = x+r-2\) and \(y = 2-x\).
By setting up these new limits properly, the newly written integral \( \int_{r-2}^{2} \int_{x+r-2}^{2-x} f(x, y) \, dy \, dx \) maintains the same region as the original setting.

Reversing order often requires expressing the boundary equations in terms of the other variable, ensuring they fit within the natural constraints of the region. This perspective not only aids in computing integrals but also deepens your understanding of how the mathematical function behaves over different domains.

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Most popular questions from this chapter

Concern the center of mass, the point at which the mass of a solid body in motion can be considered to be concentrated. If the object has density \(\rho(x, y, z)\) at the point \((x, y, z)\) and occupies a region \(W,\) then the coordinates \((\bar{x}, \bar{y}, \bar{z})\) of the center of mass are given by $$\bar{x}=\frac{1}{m} \int_{W} x \rho d V \quad \bar{y}=\frac{1}{m} \int_{W} y \rho d V \quad \bar{z}=\frac{1}{m} \int_{W} z \rho d V$$ where \(m=\int_{W} \rho d V\) is the total mass of the body. A solid is bounded below by the square \(z=0,0 \leq x \leq\) \(1,0 \leq y \leq 1\) and above by the surface \(z=x+y+1.\) Find the total mass and the coordinates of the center of mass if the density is \(1 \mathrm{gm} / \mathrm{cm}^{3}\) and \(x, y, z\) are measured in centimeters.

Are the statements in Problems \(81-88\) true or false? Give reasons for your answer. If \(f(x, y)=g(x) \cdot h(y),\) where \(g\) and \(h\) are singlevariable functions, then $$\int_{a}^{b} \int_{c}^{d} f d y d x=\left(\int_{a}^{b} g(x) d x\right) \cdot\left(\int_{c}^{d} h(y) d y\right)$$

Explain what is wrong with the statement. If \(R\) is the region \(x^{2}+y^{2} \leq 4,\) then \(\int_{R}\left(x^{2}+y^{2}\right) d A=\) \(\int_{0}^{2 \pi} \int_{0}^{2} r^{2} d r d \theta\).

Are the statements true or false? Give reasons for your answer. If \(f>g\) at all points in the solid region \(W,\) then \(\int_{W} f d V>\int_{W} g d V.\)

Are the statements true or false? Give reasons for your answer. Both double and triple integrals can be used to compute volume.

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