Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Evaluate $\int_{R} \int f(x, y) d A$ for $R$ and $f$ as given.(a) $f(x, y)=1,$(b) $f(x, y)=2 x+3 y-4$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

Campbell University

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

04:14

03:10

Evaluate $\int_{R} \int f(…

03:02

04:56

02:38

02:11

02:45

03:16

02:52

Find $$\int_{0}^{2} f(x, y…

01:26

Calculate $F(4)$ given tha…

02:07

Given that $\int_{-1}^{2} …

01:47

Evaluate the integral usin…

for this problem we are asked to evaluate the double integral of f X Y for R and F. As given where r is the region in between the lines. Now, I'll sort of outline what our actual region are is. But we have the lines here, Y equals four, X Y equals two x and Y equals x squared. Now, what we're going to have to do here Is essentially split everything apart into two different integral Depending on if we are to the left or right of x equals two. So to begin, we're trying to find the integral across that region of just one. So that will be the integral for X between zero and 2 For why? Between two x and four x dy dx. So we'd have that this would be well integrating dy would just give us why evaluated from four x or from two X to four X. We have four x minus two X. Or just integrating two X dx and integral of two X is going to be two X squared over two or just X squared Evaluated from 0-2. Which will give us a result of four. Then For the other portion we have the integral from 2-4 of the integral from X squared Up to four x dy dx. So this will then be the integral from 2 to 4 of four, X minus x four x minus x squared dx. It will then become four X squared over two or two X squared minus X cubed over three, Evaluated from 2 to 4, which will give us a final result then of 16/3. So overall our integral over the region is going to be Well, let's see here at 16/3 plus four. So that would be equal to 28/3. Then for part B we're asked to integrate two X plus three, Y minus four over that region. So we'll go through with the same order of integration. First we're integrating over Why? Then we're integrating over X integral from 0 to 2 integral from two X to four X. Now it's two X plus three Y two X plus three, Y minus four dy dx. So we'll have then the integral from 0 to 2 of two, X y plus three, Y squared over two minus four Y evaluated from two X up to four X dx. Which we can simplify down after evaluating from end to end into the form 22 X squared minus eight x DX. So now we're integrating that. So we'd have 22 x cubed over three minus eight X squared over two. So we can write this as minus four X squared Evaluated from 0 to 2, which will then give us the result 128 over three. So that is only half. We still need to evaluate the other portion of the integral. So for the other portion we'd have the integral from 2 to 4 of the integral From X squared up to four X. Let me just double check. Yeah, it's just X squared 24 x of again two X plus three y minus four dy dx. So we know that the indefinite integral of the interior is going to end up being the same thing as up here. This two X Y plus three Y squared minus four or up one second here. Just to be careful. I don't believe I miss evaluated this. But that should have been minus four. Why not? Four y squared there. But we would have that our first indefinite integral is going to be that. But when we evaluate from endpoint to endpoint, we'll have instead That the interior integral will be or that the result rather will be the integral of 36, x squared -2 x cubed -3 x. to the power of 4/2 minus 16 X Evaluated from zero. or excuse me not evaluated from anything just DX Then we're integrating this over X. So the result that we'll find then will be well 36 x cubed over three minus two X. The power 4/4. So X power 4/2 minus three X to the power of 5/2 times five. So that's three X power 5/10 minus 16 X squared over two. So that's minus X or minus eight X squared Evaluated from 2- four. When we evaluate that out, we'll find that the result will be 792/5. So the double integral over our F of x y d a. Will it end up being 1 28/3 plus 7 92 over five. Which gives us the result, then of 3016 over 15.

View More Answers From This Book

Find Another Textbook

01:53

Evaluate the given integral and check your answer.$$\int\left(\frac{5}{x…

01:18

Determine the region $R$ determined by the given double integral.$$\int_…

Euler's theorem states that for smooth homogenous functions of degree $…

Evaluate the given integral.$$\int 4^{2 x} d x$$

00:54

01:41

00:53

A function is said to be homogeneous of degree $n$ if $f(\gamma x, \gamma y)…

01:02

Solve the given differential equation.$$\frac{d y}{d x}=\frac{2 x^{3}}{y…

01:01

Solve the given differential equation.$$\frac{d y}{d x}=\frac{4 e^{x}}{y…

(a) determine a definite integral that will determine the area of the region…