Question

6={(x,y,z)[05x5,05 ys1,05=521–ya 11. Use a triple integral to find the volume of the region: E={( You must evaluate this integral by hand and show all the steps that lead to your answer, including all substitutions needed. You may not use a table of integrals or technology to evaluate this integral. Give the exact answer.

Answer 1

The volume of any region E in cartesian reo-ordinates (x, y, z) the formula is given by V SSS azdyda SSS(4), dzayda E E The lorder of Integration dzdyda can be taken as dadydz vor dydzda and so on but is chosen raccording the bounds of x, y and 2. Here; E O2 x2 ТС {{x,y,z) losne ;O=y=1; 022 <J 1-y² We chose the voeder of Integration was dzdyda ie. the Integration first takes place with respect to z and then y and at last with respect to a. 7 Volume V SSS a dzayda 든 Tel4 TY/4 si-ya 11 SS is 1-4² 1- dzdy dhe SS({ ade), ayax 2-0 Z=0 y=0 Z-0 x=0 y=0 T\/4 TY/4 s s [z z] dyda Il lui-ge-o) ay de Z=0 26=0 y=0 x=0 y=0

T/4 1 Va 51-g² dy dy da we introduce a new variable t such that y= - cost differentiating booth sides we get ) dy = - sint. at Also, the limits change y=0 va y = cost cost=0 * t = π 2 and y=1 y = cost → cost =1 * t=0 Tula va Ji-cost. (- sint) &t. da dire) cu 950 += 7/2 We know b a f (t) at -ffet) at cand l-cost = sint a b T1/4 > TY2 V = - I sint! - sintat da -0 t=o Remember sint = I sint! but since ost <T/ and sint » , more precisely os sintal > Isint/= sint

T44 T/2 Hence; va ss sinit.at da ; we use To solve the Integral int I-coset) = 2 sin't the identity >> sunt = I - costat) 2 T\/4 T/2 -55 ss ( - cos2t) dt da =0 t=0 TV4 v T\/ [t- sinat da cosmt at = sin(me) tzo Tu/4 ㅋ V= I 2 s [05.-0) -(0-0)] . da 250 Thu T1/4 < = s te da ł.da T\/4 π da 21 Hence; 2 V ТС 4 x I 16 Thus, volume of E = TC alto 16

Thus, volume of region E

$\small =\frac{\pi^{2}}{16} \\\\ =0.616850275$ ( in decimals )

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