/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Express the integral as an equiv... [FREE SOLUTION] | 91Ó°ÊÓ

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Express the integral as an equivalent integral with the order of integration reversed. $$ \int_{0}^{2} \int_{0}^{\sqrt{x}} f(x, y) d y d x $$

Short Answer

Expert verified
Reversed integral is \(\int_{0}^{\sqrt{2}} \int_{y^2}^{2} f(x, y) \, dx \, dy \).

Step by step solution

01

Understand the Region of Integration

The original double integral \( \int_{0}^{2} \int_{0}^{\sqrt{x}} f(x, y) \, dy \, dx \) represents an area in the xy-plane. For each \(x\) from 0 to 2, \(y\) ranges from 0 to \(\sqrt{x}\). This describes a region bounded by the curves \(y = 0\), \(x = 2\), and \(y = \sqrt{x}\).
02

Sketch the Region

To visualize the integration region, plot the functions. Draw the line \(y = 0\) (x-axis), the vertical line \(x = 2\), and the curve \(y = \sqrt{x}\). This region is above the x-axis and below the curve \(y = \sqrt{x}\) from \(x = 0\) to \(x = 2\).
03

Determine New Limits of Integration

To reverse the order of integration, we need to express the region in terms of \(y\) first. The values of \(y\) in the region go from 0 (bottom) to \(\sqrt{2}\) (top, since \(\sqrt{2}\) is the maximum value of \(y = \sqrt{x}\) when \(x = 2\)). For each fixed \(y\), \(x\) will range from \(y^2\) (since \(y = \sqrt{x}\) implies \(x = y^2\)) to 2.
04

Rewrite the Integral with New Limits

Rewrite the integral to reflect the new limits resulting from reversing the order of integration: \( \int_{0}^{\sqrt{2}} \int_{y^2}^{2} f(x, y) \, dx \, dy \).

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Key Concepts

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

Double Integrals
Double integrals, much like single integrals, deal with areas under surfaces. However, they do so in two-dimensional spaces. In essence, a double integral allows you to take the integral of a function over a region in the xy-plane. Instead of summing up infinitesimally small slices along one axis (as in single integrals), with double integrals, we sum up over a coordination grid, considering slices through a region. This makes double integrals crucial in computing areas, volumes, and other quantities spread over a plane. Typically denoted by \( \int \int \) or \( \iint \), these integrals are written with two integration symbols followed by the variables of integration. The function to be integrated is written in the middle. Understanding how to set and manipulate these integrals is key. Changing the order of integration, as in our exercise, shows how flexible they can be when computing specific regions, even if the process seems complex at first.
Integration Limits
Integration limits define the boundaries over which you are integrating. Just like fences create borders around a property, integration limits help in identifying the region we're looking at within double integrals. When dealing with double integrals, the order of integration is crucial. That determines how the limits are set. Given \( \int_{a}^{b} \int_{g(x)}^{h(x)} f(x,y) dy dx \), here the outer limits are for \(x\) while the inner limits are for \(y\). However, one can switch the order, requiring a re-definition of these limits.Reversing the order of integration involves expressing the integration region first in terms of \( y \). This requires finding how \( x \) changes in response to \( y \), something clearly explained via sketching or analysis. For example, when reversing \( \int_{0}^{2} \int_{0}^{\sqrt{x}} f(x, y) \, dy \, dx \) as \( \int_{0}^{\sqrt{2}} \int_{y^2}^{2} f(x, y) \, dx \, dy \), the expressions \(y^2\) and 2 signal a switch in viewing the problem.
Region of Integration
The region of integration is the specific section of the xy-plane over which we're integrating our function. In practice, this is visualized as a shape formed by bounding curves.Determining this region involves understanding graphs and curves that outline the boundaries. Here, the region was initially bounded by the curve \(y = \sqrt{x}\), the x-axis \(y = 0\), and the vertical line \(x = 2\). Upon reversing the order of integration, understanding this boundary allows redefining the region with new limits. To visualize, one can sketch these graphs. Starting with drawing each boundary line or curve helps you see where they intersect, effectively shading out the area of interest. Recognizing that the bounding curves dictate your starting and ending points is vital. With curves like \(y = \sqrt{x}\), mathematical conversion, such as \(x = y^2\), aids in re-setting integrals when reversing limits, ensuring the integration encompasses the entire and only the desired region.

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Most popular questions from this chapter

Use the transformation \(u=x+y, v=x-y\) to find $$ \iint_{R}(x-y) e^{x^{2}-y^{2}} d A $$ over the rectangular region \(R\) enclosed by the lines \(x+y=0, x+y=1, x-y=1, x-y=4\)

These exercises reference the Theorem of Pappus: If \(R\) is a bounded plane region and \(L\) is a line that lies in the plane of \(R\) such that \(R\) is entirely on one side of \(L,\) then the volume of the solid formed by revolving \(R\) about \(L\) is given by $$\text {volume}=(\text {area of } R) \cdot\left(\begin{array}{c}{\text {distance} \text {traveled}} \\ {\text {by the centroid}}\end{array}\right)$$ Use the Theorem of Pappus and the fact that the area of an ellipse with semiaxes \(a\) and \(b\) is \(\pi a b\) to find the volume of the elliptical torus generated by revolving the ellipse $$\frac{(x-k)^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ about the \(y\) -axis. Assume that \(k>a\)

Suppose that the density at a point in a gaseous spherical star is modeled by the formula $$\delta=\delta_{0} e^{-(\rho / R)^{3}}$$ where \(\delta_{0}\) is a positive constant, \(R\) is the radius of the star, and \(\rho\) is the distance from the point to the star's center. Find the mass of the star.

The tendency of a lamina to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If a lamina occupies a region \(R\) of the \(x y\) -plane, and if its density function \(\delta(x, y)\) is continuous on \(R,\) then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y},\) and \(I_{z},\) respectively, and are defined by $$\begin{aligned} I_{x} &=\iint_{R} y^{2} \delta(x, y) d A, \quad I_{y}=\iint_{R} x^{2} \delta(x, y) d A \\ I_{z} &=\iint_{R}\left(x^{2}+y^{2}\right) \delta(x, y) d A \end{aligned}$$ Use these definitions in Exercises 45 and 46 Consider the rectangular lamina that occupies the region described by the inequalities \(0 \leq x \leq a\) and \(0 \leq y \leq b\). Assuming that the lamina has constant density \(\delta,\) show that $$I_{x}=\frac{\delta a b^{3}}{3}, \quad I_{y}=\frac{\delta a^{3} b}{3}, \quad I_{z}=\frac{\delta a b\left(a^{2}+b^{2}\right)}{3}$$

Evaluate the iterated integral. $$ \int_{0}^{\pi / 2} \int_{0}^{a \sin \theta} r^{2} d r d \theta $$

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