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In Exercises \(9-22,\) change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}}\left(x^{2}+y^{2}\right) d x d y$$

Short Answer

Expert verified
The evaluated integral is \( \frac{\pi}{8} \).

Step by step solution

01

Understand the Original Integral

The given Cartesian integral is \( \int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}}(x^{2}+y^{2}) \, dx \, dy \), which represents a region of integration in the Cartesian plane. 'x' varies from \(0\) to \(\sqrt{1-y^2}\), and 'y' from \(0\) to 1.
02

Change to Polar Coordinates

In polar coordinates, \( x = r\cos\theta \) and \( y = r\sin\theta \) and \( x^2 + y^2 = r^2 \). The Jacobian of the transformation from Cartesian to polar coordinates is \( r \). The double integral becomes \( \int \int r^3 \, dr \, d\theta \).
03

Identify the Region in Polar Coordinates

The region described by \( x = 0 \) to \( \sqrt{1-y^2} \) and \( y = 0 \) to 1 translates to a quarter-circle in polar coordinates. This is because \( x^2+y^2 \leq 1 \) for \(y = 0\) to \(1\), and aligns with 0 to \( \frac{\pi}{2} \) in \(\theta\). Thus \(0 \leq r \leq 1\), \(0 \leq \theta \leq \frac{\pi}{2} \).
04

Set Up the Polar Integral

Based on the region, the integral becomes:\[\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} r^3 \, dr \, d\theta\]
05

Evaluate the Inner Integral

First, integrate with respect to \( r \):\[\int_{0}^{1} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{1} = \frac{1}{4}\]
06

Evaluate the Outer Integral

Next, integrate the result with respect to \( \theta \):\[\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \, d\theta = \frac{1}{4} \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{8}\]
07

Final Answer

Thus, the value of the original Cartesian integral, converted to polar coordinates and evaluated, is \( \frac{\pi}{8} \).

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Key Concepts

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

Cartesian coordinates
Cartesian coordinates form a foundational system used widely in mathematics to denote geometric shapes and spaces. This system assigns a pair of numbers, typically represented as \((x, y)\), to a point on a plane. Each coordinate represents a distance from the specified origins: the x-coordinate is the distance along the horizontal axis, while the y-coordinate is the distance along the vertical axis.

Using Cartesian coordinates, geometric and algebraic operations become quite intuitive. For instance, they allow for the easy calculation of distances between points, slopes of lines, and areas of geometric regions using simple equations. Cartesians lie at the heart of calculus as they enable the representation of regions through mathematical boundaries, like the integral in the initial problem.

Understanding how to work within the Cartesian system is crucial, especially as it lays the groundwork for moving into more complex coordinate spaces, such as polar coordinates.
integral transformation
An integral transformation involves converting an integral from one coordinate system to another, simplifying the calculation process. This transformation often switches Cartesian coordinates to polar coordinates. This is particularly useful when dealing with circular or radial symmetry in a problem.

In the original exercise, the function inside the integral, \(x^2 + y^2\), transformed under polar coordinates simplifies greatly, since it becomes \(r^2\). This transformation takes advantage of the polar parameters \((r, \theta)\), where \(r\) is the radius — the distance from the origin — and \(\theta\) is the angle from the positive x-axis.

Transformations are not merely algebraic steps; they reflect a shift in perspective. By switching terms into polar coordinates, the calculus often becomes more manageable, leading to straightforward integration paths.
region of integration
The concept of the region of integration is fundamental when evaluating double integrals. It defines the area over which the function is integrated. In the Cartesian setup provided, the region initially covered areas restricted by the expressions \(0 \leq x \leq \sqrt{1-y^2}\) and \(0 \leq y \leq 1\).

When we shift to polar coordinates, identifying the new region is crucial. The Cartesian limits turn into a quarter-circle defined by limits \(0 \leq \theta \leq \frac{\pi}{2}\) and \(0 \leq r \leq 1\). This simplifies where the integral is evaluated, aligning neatly with radial symmetry common in polar coordinates.

Visualizing both the Cartesian and polar regions helps deeply understand the symmetry and boundaries involved. Accurately mapping this region ensures the integration process remains correct and precise.
Jacobian determinant
The Jacobian determinant plays a significant role in transformations, indicating how a small change in the input space (like a Cartesian pair) results in a change in the output space (such as polar coordinates). Essentially, it scales the function to adjust for the distortion between coordinate planes during transformation.

In our example, converting the integral from Cartesian to polar requires introducing the Jacobian \(r\), which arises from the derivative calculations linking \(x\) and \(y\) to \(r\) and \(\theta\). Thus, the differential area element transforms from \(dx \, dy\) to \(r \, dr \, d\theta\), correcting the volume for the stretched or shrunk distances between coordinate systems.

Comprehending the Jacobian is integral for accurate coordinate transformations, ensuring that the results of integrations are cosistent and reflect the true dimensions of the region integrated over.

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Most popular questions from this chapter

How would you evaluate the double integral of a continuous function \(f(x, y)\) over the region \(R\) in the \(x y\) -plane enclosed by the triangle with vertices \((0,1),(2,0),\) and (1,2)\(?\) Give reasons for your answer.

The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals. $$\int_{\pi / 6}^{\pi / 3} \int_{\csc \phi}^{2 \csc \phi} \int_{0}^{2 \pi} \rho^{2} \sin \phi d \theta d \rho d \phi$$

Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder \(x^{2}+y^{2}=4,\) and the plane \(z+y=3\)

Show that the centroid of a solid right circular cone is one-fourth of the way from the base to the vertex. (In general, the centroid of a solid cone or pyramid is one-fourth of the way from the centroid of the base to the vertex.)

approximate the double integral of \(f(x, y)\) over the region \(R\) partitioned by the given vertical lines \(x=a\) and horizontal lines \(y=c .\) In each subrectangle, use \(\left(x_{k}, y_{k}\right)\) as indicated for your approximation. $$\iint_{R} f(x, y) d A \approx \sum_{k=1}^{n} f\left(x_{k}, y_{k}\right) \Delta A_{k}$$ \(f(x, y)=x+y\) over the region \(R\) bounded above by the semicircle \(y=\sqrt{1-x^{2}}\) and below by the \(x\) -axis, using the partition \(x=-1,-1 / 2,0,1 / 4,1 / 2,1\) and \(y=0,1 / 2,1\) with \(\left(x_{k}, y_{k}\right)\) the lower left corner in the \(k\) th subrectangle (provided the subrectangle lies within \(R\) )

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