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Double Integrals & Evaluation by Iterated Integrals

Chapter 15 /  Lesson 4   Transcript
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  • 0:04 Iterating
  • 0:24 Iteration Integration
  • 2:20 A Special Case
  • 3:14 Lesson Summary
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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we show how to evaluate a double integral using iterative integration. A special case is also presented which simplifies the calculations.

Iterating

The word ''iteration'' means to do something repeatedly. This word is often used in math. In this lesson, we evaluate a double integral by using iteration. First, we'll do a single integral on one of the variables and then repeat with a single integral on the other variable. Rather than repeat this information, let's do an example.

Iteration Integration

Imagine needing to find the solution to the double integral


int_x=2,3;y=-1,4;yx^2_dxdy


We have two variables, x and y. Let's integrate on x first. To be clear, we can use parentheses:


int_y=-1,4;(int_x=2,3;yx^2_dx)dy


The integral inside the parentheses is called the inner integral. We need to evaluate the integral of yx2 from x = 2 to x = 3.

This is an integration on x. What about the y appearing in yx2? In this integration on x, we treat y as if it were a constant. So, yx2 is like cx2 where c is a constant.

The integration of cx2 is cx3/3. So, when we integrate on x, the integral of yx2 is yx3/3.

Evaluating with the upper and lower limits:


int_x=2,3_yx^2dx=yx^3/3_2_3=y/3(3^3-2^3)


Simplifying, we get 19y/3 as the answer for the inner integral. Next, we do the outer integral. We evaluate:


int_y=1,4_19y/3+dy


Do you see how the integral within the parentheses has been replaced with 19y/3?

Now, we iterate (i.e, repeat) by integrating over y:


int_y=1,4_19y/3_dy=19y^2/6_1,4=19(4^2-(-1)^2)/6


Simplifying, we get 47.5 as the final answer.

The double integral was computed using an ''iterated integral.'' We started with an integration on x. What if we had started with an integration on y?

Then, the double integral would be written as:


int_x=2,3(int_y=-1,4_yx^2_dy)dx


Do you see how the work is clarified by using parentheses and by labeling the lower limit of the integrals?

The integration on y becomes:


int_y=-1,4_yx^2_dy=x^2y^2/2_-1,4


Now, the variable is y while x is the constant. Continuing:


=x^2/2(4^2-(-1)^2=x^2/2(16-1)=15x^2/2


Now, the outer integral is over x:


int_x=2,3_15x^2/2_dx=15/6_x^3_2,3=5/2(3^3-2^3)


Do you see how 15 divided by 3 produced 5?

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