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Use polar coordinates to combine the sum $$\int_{1 / \sqrt{2}}^{1} \int_{\sqrt{1-x^{2}}}^{x} x y d x d x+\int_{1}^{\sqrt{2}} \int_{0}^{x} x y d y d x+\int_{\sqrt{2}}^{2} \int_{0}^{\sqrt{4-x^{2}}} x y d y d x$$ into one double integral. Then evaluate the double integral.

Short Answer

Expert verified
The integral evaluates to \( \frac{9}{16} \).

Step by step solution

01

Sketch the Region

First, let's understand the region of integration by analyzing each part of the piecewise double integrals provided. Sketching a diagram is helpful to see the constraints of each integral in the xy-plane and later in the polar coordinate system.
02

Convert Limits to Polar Coordinates

Next, convert the Cartesian limits of integration to polar coordinates. Recall the transformations: \(x = r \cos \theta\) and \(y = r \sin \theta\). The regions described in the original integrals correspond to portions of a circle defined by \(1/\sqrt{2} \leq r \leq 2\) and appropriate angular limits.
03

Combine the Integrals

Notice each of the regions described essentially makes up the sector of a circle. Using polar coordinates, you can simplify the integrals into a single integral over a single region. Sum them up by defining a single integral over all angles \(\theta\) from \(\pi/4\) to \(\pi/2\) for the region \( 1/\sqrt{2} \leq r \leq 2 \), which encloses the complete region.
04

Express the Integral in Polar Coordinates

Use the conversion formulas and express the double integral in polar coordinates as \[ \int_{\pi/4}^{\pi/2} \int_{1/\sqrt{2}}^{2} r^2 \cos(\theta) \sin(\theta) \, dr \, d\theta \]. Notice that \(xy\) converts to \(r^2 \cos(\theta) \sin(\theta) \) after using \(x=r\cos\theta\) and \(y=r\sin\theta\) and factoring the \(r\) from \(dxdy\).
05

Evaluate the Integral

Evaluate the inner integral with respect to \(r\), using \[\int r^3 \, dr\] over the limits from \(1/\sqrt{2}\) to \(2\), and then the outer integral with respect to \(\theta\). Simplify using trigonometric identities if necessary, such as \(2\cos\theta\sin\theta = \sin(2\theta)\).
06

Calculate the Definite Integral

Complete the calculations for both integrals and substitute the values back. Following through with the correct integration details, you should find the numerical answer. Be careful to compute areas using exact values and use an antiderivative for both integrations.

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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 an essential concept in multivariable calculus. They allow you to calculate the volume under a surface or over a given region in a plane. Integrating once calculates the area under a curve, while integrating twice considers a region in two dimensions:
  • The double integral symbol is written as \(\int\int\) and usually includes two sets of limits.
  • The inner integral is evaluated first, holding one variable constant, while the outer integral sums up these results across the second variable.
Understanding the geometry of the region by sketching it or visualizing it is crucial. This way, you can see how the limits of integration form a complete picture. For example, in the original exercise, the combined double integral represents the sum of several smaller regions, each defined by specific limits. By converting similar pieces into a single piece, these can be analyzed together, simplifying the entire process.
Conversion to Polar Form
When dealing with integrals over circular or symmetric regions, converting Cartesian coordinates to polar coordinates can simplify matters. Polar coordinates use a radius and angle to describe points in the plane:
  • Cartesian coordinates \((x, y)\) translate to polar as \(x = r \cos \theta\) and \(y = r \sin \theta\).
  • The Jacobian for converting dxdy to polar is \(r\), derived from \(r^2 = x^2 + y^2\).
This conversion is especially handy when the region of integration involves curves like circles or parts of circular areas. In the given exercise, transforming the integral boundaries to polar form helps align the regions into a continuous sector of a circle. This approach blends distinct parts of the region into a seamless integral by altering the variables. Handling these polar conversions proficiently enables you to deal with more complex regions by simplifying the limits into a straightforward radial and angular form.
Integration Limits
Integration limits specify the boundaries of the area over which you're integrating. In polar coordinates, these limits often involve angles and radii rather than straight lines:
  • Lower and upper limits for \(r\) define how far from the origin the region extends.
  • \(\theta\) limits describe sections of the circle covered by the area.
In our case, converting from Cartesian to polar limits makes it easier to combine different pieces of the original problem into one integral. For this exercise, the limits became \(1/\sqrt{2} \leq r \leq 2\) for the radius and \(\pi/4\) to \(\pi/2\) for the angle.
By getting the integration limits correct, you effectively map the region needing evaluation. Changes in limits correspond to the shape and dimensionality of the area; understanding how they align not only improves calculation accuracy but also enhances visualization of the problem. Incorporating exact ranges ensures your integral spans the whole intended region, simplifying further evaluation.

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Most popular questions from this chapter

Evaluate the iterated integral by converting to polar coordinates. \(\int_{-3}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sin \left(x^{2}+y^{2}\right) d y d x\)

Evaluate the iterated integral by converting to polar coordinates. \(\int_{0}^{2} \int_{0}^{\sqrt{2 x-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x\)

Evaluate the given integral by changing to polar coordinates. \(\iint_{R} \sin \left(x^{2}+y^{2}\right) d A,\) where \(R\) is the region in the first quadrant between the circles with center the origin and radii 1 and 3

The latitude and longitude of a point \(P\) in the Northern Hemisphere are related to spherical coordinates \(\rho, \theta, \phi\) as follows. We take the origin to be the center of the earth and the positive \(z\) -axis to pass through the North Pole. The positive \(x\) -axis passes through the point where the prime meridian (the meridian through Greenwich, England) intersects the equator. Then the latitude of \(P\) is \(\alpha=90^{\circ}-\phi^{\circ}\) and the longitude is \(\beta=360^{\circ}-\theta^{\circ} .\) Find the great-circle distance from Los Angeles (lat. \(34.06^{\circ} \mathrm{N},\) long. \(118.25^{\circ} \mathrm{W}\) ) to Montréal (lat. \(45.50^{\circ} \mathrm{N},\) long. \(73.60^{\circ} \mathrm{W} ) .\) Take the radius of the earth to be 3960 \(\mathrm{mi}\) . (A great circle is the circle of intersection of a sphere and a plane through the center of the sphere.)

Use the given transformation to evaluate the integral. $$\begin{array}{l}{\iint_{R} x y d A, \text { where } R \text { is the region in the first quadrant }} \\ {\text { bounded by the lines } y=x \text { and } y=3 x \text { and the hyperbolas }} \\ {x y=1, x y=3 ; \quad x=u / v, y=v}\end{array}$$

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