mth 254 – mr. simonds’ class - portland community...

MTH 254 – Mr. Simonds’ class

R e c t a n g u l a r D o u b l e I n t e g r a l s : S e c t i o n s 1 2 . 1 - 1 2 . 3 | 1

Suppose that we wanted to determine the volume of the solid shown in Figure 1. The lateral sides of this solid are formed by the planes 0x = , 0y = , and

6x y+ = . The bottom of the solid lies on the xy‐plane and the top of the solid lies

along the surface 3 25z x y y= − + In MTH 252, we estimated the area of planar regions by superimposing rectangles over the region and adding up the areas of these rectangles. We can estimate the volume of our solid in a similar way; we can superimpose rectangular parallelepipeds over the solid and add up the volumes of these boxes. In Figure 2, the y‐axis of the region the solid projects onto the xy‐plane has been subdivided into 3 intervals of equal width. Over each of these intervals a rectangle has been formed between the

boundaries 0x = and 6x y= − using the values of x at the midpoint, *jy , to determine the position

in which to draw the line segments determined by 0x = and 6x y= − . This is summarized in Table 1. Table 1

k interval *ky

rectangular borders

( x borders come from 0x = and *6 kx y= − )

1 [ ]0,2 1 0, 2, 0, 5y y x x= = = =

2 [ ]2,4 3 2, 4, 0, 3y y x x= = = =

3 [ ]4,6 5 4, 6, 0, 1y y x x= = = =

Each of the rectangular regions in Table 1 has been further partitioned into 2 rectangles of equal length – this partition happening parallel to the x‐axis. Rectangular parallelepipeds are now formed using the 6 rectangles in Figure 2 as bases. The tops of the

boxes are portions of the planes ( )* *,i jz f x y= where ( )* *,i jx y is the midpoint of rectangleij and

( ), 3 25f x y x y y= − + . The tops and bases of the boxes are shown in Figure 3 and a facsimile of the

6 boxes is shown in Figure 4. The top of the actual solid is shaded in Figure 5. Notice how each of the rectangular parallelepipeds has some volume that lies outside the actual solid while part of the actual solid does not lie inside any of the rectangular parallelepipeds. The more balanced these two discrepancies, the better our estimate.

Figure 2

Figure 3 Figure 5 Figure 4

Figure 1


2 | R e c t a n g u l a r D o u b l e I n t e g r a l s : S e c t i o n s 1 2 . 1 - 1 2 . 3

Let’s go ahead and use this scheme to estimate the volume of the solid. Let’s use Figure 6 and Table 2 to help us come up with our estimate. Table 2

,i j x‐interval y‐interval ( )* *,i jx y ( )* *,i jf x y ijAΔ

1,1

2,1

1,2

2,2

1,3

2,3

Figure 6



Let’s use Figure 7 to help us set up a double integral of form ( ),R

f x y dxdy∫∫ that finds the exact

volume of our solid. Let’s evaluate the integral by hand. In this form, the y‐limits of integration are points on the y‐axis and the x‐limits of integration are

boundary curves of form ( )x L y= and ( )x R y= .

Figure 7



Let’s use Figure 8 to help us set up a double integral of form ( ),R

f x y dydx∫∫ that finds the exact

volume of our solid. Let’s evaluate the integral on our calculators. In this form, the x‐limits of integration are points on the x‐axis and the y‐limits of integration are

boundary curves of form ( )y B x= and ( )y T x= .

Figure 8



Consider ( )2 5 2

3 2

xy e y dy dx−

−∫ ∫ . Let’s illustrate the region of integration on Figure 9, paying close

attention to what are axis limits and what are boundary curve limits. Let’s calculate the value of the integral by hand. Let’s illustrate on Figure 10 what it is, exactly, that we just calculated.

Figure 9

Figure 10



Let’s use our calculator to find the exact value of 26 36 2

6 0

xx y dy dx

−

−∫ ∫ . Let’s clearly label the region of

integration on Figure 11.

Let’s use our calculator to find the exact value of 26 36 2

6 0

xx y dy dx

−

−∫ ∫ after first reversing the order of

integration. Let’s clearly label the region of integration on Figure 12.

Figure 12

Prep Work

Figure 11



The curves xy 2= and 2

2xy = are shown in figures 14 and

15. In Figure 13, these curves have been laterally extended up to the plane 2=z . We are going to set up both of the double integrals that calculate this volume. We are also going to evaluate each double integral (and hopefully get the same value both times! ☺)

Figure 14

Figure 13

Figure 14



On this page we are going to find both double integrals that calculate the volume of the tetrahedron formed in the first octant by the plane in Figure 15. We will then use technology to calculate each double integral and confirm that we get the same value twice. ☺

Figure 16

The Tetrahedron in question

Figure 15

Figure 17



The top of the solid outlined in Figure 118 is the plane xz = . The curved boundary in the xy‐plane is

the function ( )lny x= . We are going to set up both

double integrals that calculate the volume of the solid. We are (mercifully) going to evaluate the double integrals using technology. ☺

Figure 18

Figure 19

Figure 20



Finally, we are going to calculate 24 2

0 /2

y

xe dy dx−∫ ∫ after first reversing the order of integration.

Figure 21

Figure 22

mth 254 – mr. simonds’ class - portland community...

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