/*! 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 7 Evaluate the following iterated ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 7

Evaluate the following iterated integrals. $$\int_{1}^{3} \int_{0}^{2} x^{2} y d x d y$$

Short Answer

Expert verified
Question: Evaluate the iterated integral $$\int_{1}^{3} \int_{0}^{2} x^{2} y d x d y.$$ Answer: The value of the given iterated integral is $$\int_{1}^{3} \int_{0}^{2} x^{2} y d x d y = \frac{28}{3}.$$

Step by step solution

01

Integrate with respect to x

Integrate the inner integral first, holding y constant as we consider x. The function inside the integral is x^2 * y. When we take the antiderivative with respect to x, we obtain: $$\int x^2 y dx = \frac{x^3 y}{3} + C_1(y)$$ We intentionally add the constant, \(C_1(y)\), as a function of y to emphasize that we're only integrating with respect to x at this point.
02

Evaluate the x-limits

We now evaluate the antiderivative from step 1 over the x-interval given, which is from 0 to 2: $$\frac{2^3 y}{3} - \frac{1^3 y}{3} = \frac{8y - y}{3} = \frac{7y}{3}$$
03

Integrate with respect to y

Now that we have simplified the expression after the x-integration, we can integrate this with respect to y: $$\int \frac{7y}{3} dy = \frac{7}{6}y^2 + C_2$$ Here, \(C_2\) is another constant term which will vanish as we evaluate the integral upper and lower bounds in the next step.
04

Evaluate the y-limits

Lastly, we evaluate the y-interval from 1 to 3: $$\frac{7}{6}(3^2) - \frac{7}{6}(1^2) = \frac{7}{6}(9 - 1) = \frac{7}{6}(8) = \frac{28}{3}$$ So, the value of the given iterated integral is: $$\int_{1}^{3} \int_{0}^{2} x^{2} y d x d y = \frac{28}{3}$$

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 91影视!

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

A thin (one-dimensional) wire of constant density is bent into the shape of a semicircle of radius \(a\). Find the location of its center of mass. (Hint: Treat the wire as a thin halfannulus with width \(\Delta a,\) and then let \(\Delta a \rightarrow 0\).)

General volume formulas Use integration to find the volume of the following solids. In each case, choose a convenient coordinate system, find equations for the bounding surfaces, set up a triple integral, and evaluate the integral. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Ellipsoid Find the volume of a solid ellipsoid with axes of length \(2 a, 2 b,\) and \(2 c\).

Use spherical coordinates to find the volume of the following solids. The solid bounded by the cylinders \(r=1\) and \(r=2,\) and the cones \(\varphi=\pi / 6\) and \(\varphi=\pi / 3\)

An important integral in statistics associated with the normal distribution is \(I=\int_{-\infty}^{\infty} e^{-x^{2}} d x .\) It is evaluated in the following steps. a. Assume that $$\begin{aligned} I^{2} &=\left(\int_{-\infty}^{\infty} e^{-x^{2}} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2}} d y\right) \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} d x d y \end{aligned}$$ where we have chosen the variables of integration to be \(x\) and \(y\) and then written the product as an iterated integral. Evaluate this integral in polar coordinates and show that \(I=\sqrt{\pi} .\) Why is the solution \(I=-\sqrt{\pi}\) rejected? b. Evaluate \(\int_{0}^{\infty} e^{-x^{2}} d x, \int_{0}^{\infty} x e^{-x^{2}} d x,\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\) (using part (a) if needed).

Improper integrals arise in polar coordinates when the radial coordinate \(r\) becomes arbitrarily large. Under certain conditions, these integrals are treated in the usual way: $$\int_{\alpha}^{\beta} \int_{a}^{\infty} f(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} f(r, \theta) r d r d \theta$$ Use this technique to evaluate the following integrals. $$\int_{0}^{\pi / 2} \int_{1}^{\infty} \frac{\cos \theta}{r^{3}} r d r d \theta$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.