K tDonkey.tnteffffsguerRectmg6s

YaYTfg

myR Ca b x E d

tyT Bi fCxi yEaxedm i

j

The volumeunder the surface and above R inXy plane is

bin EI IE ftXi't y AAM n

V f fCx.y d

If fix y DA f fab fanDdxdry16 fo fo Cxtyfdxdy

R 6,13 6 Diterated integral

this is a doubleintegral

Jo'sCxtyPl dy

I typ y dy

T2City t y41

a 6 D D

Fubini's Theorem If f is cont on

R Cx y a Eb c Eyed then

f fundA fdofabflxiydxdy fafodflx.isdydx

If 2 fix y 20 on R then the integralcan represent a volume

A this holds if f is bounded Quite andhas a finite number of discontinuities

If fCxc4 can be factored flex y gcx hey

then gcxihlxldA fbagcxsdxf.dkCuddy

Where R a b x c d

Rationale in fed gxthly dxdy hey isconstant WRT X So this becomes fedhly bagaddxdy

But Ja9CHdx is constant WRT y so we have

fabgcxldxfc.dkCuddy

4 Estimate the volume of the solid that

lies below the surface 2 It x't 3g and

above the rectangle R 1,23 10,3I EXEZ OEYE3

Use a Riemann scam with m n 2dftgeCxi 9E BXAYsubinterraestsubihtervals

for for yM DX _I dy Iz 4 2,92 a Use lower left corners

is a i 3D Xi Yii.DT.DK x vs 2T It 23 t I.EIE IZ Z 4 4

XYltx4 3y VZ15 rs

I O Z

I I1Z 15 class as2 If 375

76

10 Evaluate the doable integralby forestidentifying it as the volume of a solid

ffp 2 1 DA R CX 9 I 0 4 2 Ot y 2 4

g

e

yEn f

2 2 1 qI

I goX

a HI E 2 1

x MyVE 2 4 47 8

20 I If hfytdydxfpxdxffh fdyu haduty.dz

5

half Ethyl

t_tzhr3Chns7

Ch55Fln52tizks.fgcyxy 7dA.R _fcx.ggoexE2iHtt2tdxdyo

dydxJ

focy x54dxdy fig xy Ddydx

fyxy tzxy 4 dz

f y 257dg

yet 51,241 1 4 27