Lesson 29 MA Nick Egbert

Transcription

1 Lesson 9 MA 16 Nick Egbert Overview In this lesson we build on the previous two b complicating our domains of integration and discussing the average value of functions of two variables. Lesson So far the domains of integration which we have considered have been prett simple: rectangles and regions bounded b a function and horizontal and vertical lines. We can complicate things slightl b considering two tpes of regions. The first tpe is where is bounded above and below b functions of x. Some examples taken from Stewart s Calculus are shown below. Notice if ou follow along from left to right, there is alwas a top function and a bottom function on the boundar. Furthermore, since ranges from g 1 x to g 1, if we encounter this tpe of integral, we must set it up as fx, b gx a g 1x fx, d dx. The second tpe of region that we encounter is where is bounded to the left and right b functions of. A couple examples also from Stewart are given below. You ll notice that following along from bottom to top there is alwas a left function and a right function on the boundar. This time ou ll notice that x ranges from h 1 to h, so if we encounter a region of this tpe, we must set up our integral as fx, e h c h 1 fx, dx d. 1

2 Lesson 9 MA 16 Nick Egbert So from now on when given an integral fx, da, we will have to determine which tpe of region we have. Let s tr it in an example. Example 1. Evaluate R 8x + 7 da where R is the region bounded b x and x +. Solution. We start b drawing a picture of our domain of integration R. R x Here R is bounded above and below b functions of x, so we will want our inside integral to be d. To determine the bounds for x, we want to find the points of intersection of the line and the parabola, and we do this b setting their equations equal to one another. x x + x x x + 1x Now that we ve verified that the two curves intersect at x and x, we need to know the range of in R. The smallest can be is on the parabola, and the largest can be is on the line. So x x. Now we have enough

3 Lesson 9 MA 16 Nick Egbert information to compute the double integral. x+ 8x + 7 d dx x [ 4x + 7 x+ x ] dx [ 4xx + + 7x + 4x + 7x ] dx x + 4x + 9x + x + 14 dx x6 + x 4 + x + x + 14x 76. Sometimes our domain of integration isn t so nice. But there s a nice calculus fact that can help us out. Big fact. If can be divided into two parts 1 and so that 1 and don t overlap except for their boundaries, then fx, da fx, da + 1 fx, da. B this fact, if we are given a domain of integration that isn t one of the two nice tpes we discussed earlier, we can divide it up to make it one of the two tpes and compute each integral separatel. Example. Evaluate 4x da where is the region in the first quadrant bounded b the hperbola x 4 and the lines x, and x 4. Solution. We start b drawing a picture of. We should figure out where the curves intersect to help us complete the picture. Using x in the equation x 4, we get x 4, or x since we are in the first quadrant.

4 Lesson 9 MA 16 Nick Egbert x 4 x x 4 x 4 Notice that there is no wa to pick out a function that is alwas on top or alwas on the right, so we ll have to divide up. One wa to do this is to make a division at x. 1 x 4 Now we compute 1 4x da and 4x da separatel. For the first one, 1 4x da x x x dx 4x d dx And for the second, 4x da 4 4/x 4 16x 4x d dx 8x Now using the big fact, 4x da

5 Lesson 9 MA 16 Nick Egbert In Lesson we discussed the average value of a function of a single variable. We started with the idea of taking a discrete average: add up all the values and divide b the number of values ou have. We extended this b saing that if we have a function f on an interval [a, b], then the average value of f on [a, b] is given b f avg 1 b a b a fx dx. 1 We can extend this notion again to find the average value of a function of two variables. Given a function fx, and a region R we can define the average value of fx, over R as f avg 1 fx, da, AR R where AR denotes the area of R. If ou think about it, b a is like the onedimensional area of [a, b], so this formula isn t reall all that different from 1. Of course the simplest such region to deal with is a rectangle as in the next example. Example. Find the average value of the function fx, 6x + 4 over the rectangle R [, 4] [, 1]. Solution. Let s start b computing AR. R is the following rectangle. 1 4 So AR Moreover, x ranges from to 4 and x

6 Lesson 9 MA 16 Nick Egbert ranges from to 1. Putting this all together into the formula from, f avg x + 4 d dx 1 6x + dx 6x + dx 4 x + x Recall from Calculus I that if we have a nonnegative function f on an interval [a, b], then we can interpret the definite integral b fx dx as the are under a the curve on the given interval. We have a similar notion in functions of two variables. That is if fx, is nonnegative over a region R, then the integral fx, da R is interpreted as the volume under the surface z fx, and above R. Example 4. Find the volume under the surface z x above the triangle T whose vertices are,,, 8,, and, 7,. Solution. Since the surface z x lives in a -dimensional space, we can talk about T as living in dimensions as well. But notice that the z-coordinate of each vertex of T is, so T lives in the x-plane. Note that b the discussion above, what we re after is x da. We should draw a picture. 7 T x gx 8 x 6

7 Lesson 9 MA 16 Nick Egbert If we knew an equation for the hpotenuse of the triangle, then we would be good to go. In order to do this recall the point-slope form of a line mx x where m is the slope of the line and x 1, 1 is a particular point on the line. To calculate the slope we use m x 7 8. Then we pick a point on the line and plug it into??. 4 x x x Now, from the picture we see that ranges from to the line while x ranges from to 8. Putting this into a double integral, T x da 4 41 x+ 1 1 x x d dx 4 x x + 41 x [ 1 x 8 9 x 164 x x dx 9 dx 16 9 x 8 9 x x x x dx 9 ] x dx 7