/*! 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 11 Change the Cartesian integral in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$\int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}}\left(x^{2}+y^{2}\right) d x d y$$

Short Answer

Expert verified
The value of the polar integral is \( 2\pi \).

Step by step solution

01

Understanding the Region of Integration

The given Cartesian integral is \( \int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}}\left(x^{2}+y^{2}\right) d x d y \). We first need to understand the region it covers. The outer integral is from \( y = 0 \) to \( y = 2 \) and the inner integral is from \( x = 0 \) to \( x = \sqrt{4-y^2} \). This defines a quarter circle in the first quadrant with radius 2.
02

Convert to Polar Coordinates

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

Set Up Polar Integral

For the region defined, \( r \) ranges from \( 0 \) to \( 2 \) and \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{2} \). Substitute these limits into the polar integral form: \[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \, dr \, d\theta \].
04

Evaluate the Inner Integral

Calculate the inner integral first: \( \int_{0}^{2} r^3 \, dr \). The antiderivative of \( r^3 \) is \( \frac{r^4}{4} \). Evaluate this from 0 to 2: \[ \left[ \frac{r^4}{4} \right]_{0}^{2} = \frac{16}{4} - 0 = 4 \].
05

Evaluate the Outer Integral

Now, substitute the result of the inner integral into the outer integral: \( \int_{0}^{\frac{\pi}{2}} 4 \, d\theta \). The antiderivative of \( 4 \) with respect to \( \theta \) is \( 4\theta \). Evaluate this from 0 to \( \frac{\pi}{2} \): \[ \left[ 4\theta \right]_{0}^{\frac{\pi}{2}} = 4 \times \frac{\pi}{2} = 2\pi \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Cartesian integral
A Cartesian integral is a double integral calculated with respect to Cartesian coordinates, which are typically denoted by the variables \( x \) and \( y \). In this system, we are working on a two-dimensional plane where the x- and y-axes intersect at the origin (0,0). Cartesian integrals are often used to calculate areas under a curve or other regions defined in this coordinate system.

When setting up a Cartesian integral, it is crucial to identify the limits of integration. These limits describe the region over which the integration will be performed. For example, in the integral \( \int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}} (x^2 + y^2) \, dx \, dy \), the outer limits with respect to \( y \) span from 0 to 2, while the inner with respect to \( x \) depend on \( y \), from 0 to \( \sqrt{4-y^2} \).

This structure helps define a specific area on the plane. However, when the region is circular, converting to another system like polar coordinates can simplify the problem.
integration
Integration is a mathematical process used to find the whole by combining pieces over specified intervals. It's essentially the reverse operation of differentiation. In the context of area calculation, integration allows us to sum up infinitesimal pieces to find the area under curves or within closed regions.

When evaluating the given integral \( \int (x^2 + y^2) \, dx \, dy \), we apply definite integration over the specified limits to find the total area or volume described by the function \( x^2 + y^2 \).

This process of integration involves finding an antiderivative of the function, substituting the limits, and calculating the resulting expression. It's important to handle integration carefully, especially with transformation techniques like switching from Cartesian to polar coordinates.
polar transformation
Polar transformation is a method used to convert Cartesian coordinates \((x, y)\) into polar coordinates \((r, \theta)\). This is particularly useful for dealing with problems that have circular symmetry, such as those describing regions in circles or semicircles.

In polar coordinates:
  • \( x = r\cos(\theta) \)
  • \( y = r\sin(\theta) \)
  • Radial distance is represented by \( r \)
  • Angle from the positive x-axis is \( \theta \)

By performing this transformation, the double integral \((x^2+y^2) \, dx \, dy\) converts to \( r^3 \, dr \, d\theta \). The Jacobian, which is \( r \) for this change of variables, accounts for the stretching factor that occurs during the transformation.

Therefore, our updated polar integral for this problem becomes \( \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \, dr \, d\theta \), simplifying the computation over circular regions by using forms aligned with the geometry of the problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. a. Plot the Cartesian region of integration in the \(x y\) -plane. b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for \(r\) and \(\theta\) c. Using the results in part (b), plot the polar region of integration in the \(r \theta\) -plane. d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility. $$\int_{0}^{1} \int_{x}^{1} \frac{y}{x^{2}+y^{2}} d y d x$$

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. $$\iint_{R} f(x, y) d A=\sum_{i=1}^{n} f\left(x_{k}, y_{k}\right) \Delta A_{k}$$ \(f(x, y)=x+2 y\) over the region \(R\) inside the circle \((x-2)^{2}+(y-3)^{2}=1\) using the partition \(x=1,3 / 2,2,5 / 2\) 3 and \(y=2,5 / 2,3,7 / 2,4\) with \(\left(x_{k}, y_{k}\right)\) the center (centroid) in the \(k\) th subrectangle (provided the subrectangle lies within \(R\) )

Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry threeeighths of the way from the base toward the top, show, by transforming the appropriate integrals, that the center of mass of a solid semiellipsoid \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)+\left(z^{2} / c^{2}\right) \leq 1, \quad z \geq 0\), lies on the \(z\) -axis three-eighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.)

Integrate \(\quad f(x, y)=\) \(\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}\) over the region \(1 \leq x^{2}+y^{2} \leq e\)

Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{8} \int_{\sqrt{x}}^{2} \frac{d y d x}{y^{4}+1}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.