Exercise Qu. 12. a2 y 2 da = Qu. 18 The domain of integration: from y = x to y = x 1 3 from x = 0 to x = 1. = 1 y4 da.

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MAH MAH Eercise. Eercise. Qu. 7 B smmetr ( + 5)dA + + + 5 (re of disk with rdius ) 5. he first two terms of the integrl equl to becuse is odd function in nd is odd function in. (see lso pge 5) Qu. b da Qu. da dd ( ) d + ( ) d + ( ) ( ). ( ) d( ) d volume of the qurter clinder shown in the figure. (b ) b. Qu. 8 he domin of integrtion: from to from to dd da ( s shown) dd d u du + ( ) + 8 6. d ( ) ( ) d( ), let u, then du d

MAH MAH Qu. (see lso pge 5) Qu. Since F () f() nd G () g() on b, we hve nd intersect on the clinder +. he volume ling below nd bove is V ( )da + + 8 8 ( )da. ( )dd ] [( ) d ( ) d, let sin u, then d cos udu cos udu Alterntivel, using polr, we hve V ( + cos u) du ( + cos u + + cos u )du [ r ( r )r drdθ r ]. O I I I I, thus f()g() da f()g() dd ( f() ) G ()d d f()[g() G()]d f()g()d G() f() d f()g()d G()F(b) + G()F(). f()g() da f()g() dd ( ) b g() f() d d g()[f(b) F()] d F(b)G(b) F(b)G() F()g() d. f()g()d F(b)G(b) F()G() F()g() d.

MAH MAH Qu. sin dd. Qu. Eercise....6.8 - - da ( ) d d d ( ) (converges) (wh!!). his is the volume of the region bounded b sin, the -plne, nd the plnes,, nd. he volume is V sin dd cos d. d At this point we use the definition of bsolute vlue to split this into two quntities Qu. 5 Q + ( + )( + ) da which diverges to infinit, since Q ( + )( + da (b smmetr) ) + d + d + d V d + d + on [, ) + 9 or. + s. (see lso pge 5) 5 6

Qu. S ( + ) da other itertion: S ( + ) da + dd, ( + ) let u +, then du d u u dud ( u ) u + d ( + ) d. + dd, ( + ) let u +, then du d u u dud ( u ) u + d ( + ) d. MAH Qu. If {(,) + h, b b + k} Similrl, hus f (,)da f (,)da +h +k +h b f (,)dd [f (,b + k) f (,b)]d f( + h,b + k) f(,b + k) f( + h,b) + f(,b). +k +h b f (,)dd f( + h,b + k) f( + h,b) f(,b + k) + f(,b). f (,)da f (,)da. MAH ivide both sides of this identit b hk nd let (h,k) (,) to obtin, using the men-vlue theorem, f (,b) f (,b). hese seemingl contrdictor results re eplined b the fct tht the given double integrl is improper nd does not, in fct, eist, tht is, it does not converge. o see this, we clculte the integrl over certin subset of the squre S, nmel the tringle defined b < <, < < ( + ) da which diverges to infinit. dd, ( + ) let u +, then du d u u dud ( u u d ) d 5-5 -.8.6.....6.8 7 8

MAH MAH Eercise. Qu. Qu. + + ( + )da S r (r cos θ + r sin θ)r drdθ (cos θ + sin θ) dθ [sin θ cos θ] ( + ). 6 his is sphere centre t (,,) with rdius. In clind. coord. r + + ( ) his is clinder centre t, with rdius. In polr coord. r cos θ. B smmetr, one qurter of the required volume lies in the first octnt. ( ( ) + ). (see lso pge 5) Qu. + ln( + )da + (ln r ) r drdθ r ln r dr + r ln r [. + ] r Note tht the integrl is improper, but converges since r dr lim r ln r. r + V da cos θ cos θ r r dr ( r ) d( r ) ( r ) / cos θ + ( ) sinθ dθ dθ ( sin θ)dθ since sin θ > for θ ( cos θ)d(cos θ) + [ cos θ cos θ + [ ( ] ) ] 9 ( ). 9

MAH MAH Qu. 6 One qurter of the required volume V is shown in the figure. Qu. 7 + + ( ) his is circle centre t (,) with rdius. In polr coord. r sin θ. hus Erf() e t dt e s ds. [Erf()] e (s +t ) dsdt, S where S is the squre S {(s,t) s, t }. V da sin θ r sin θ rdrdθ sin θ 5 r5/ sin θ sin θ dθ 5 6 5. dθ B smmetr, [Erf()] 8 e (s +t ) dsdt, where {(s,t) s, t s}. In polr, we hve [Erf()] 8 sec θ ( e r ) e r r drdθ sec θ dθ ( e /cos θ ) dθ. Since cos θ, we hve e / cos θ e, so [Erf()] ( e ) dθ e Erf() e.

MAH MAH Qu. he cone + nd the sphere + ( ) + intersect where Alterntivel, let V be the volume inside the cone nd outside the sphere, then + + i.e. on the clinder In polr + ( + ) ( ). r sin θ. he volume V ling outside the cone nd inside the sphere lies on four octnts; one qurter of it is in the first octnt. o clculte V, we first clculte the volume V under the cone nd inside the clinder, i.e. V + da sin θ 9. r r drdθ sin θ dθ V V V, ( ) dda + [ ( ) + ] da ( ) da + da where V ( ) da, let ( ) da, where da dd 9 ( ) (from E.. Qu. ) V + da 9 (from bove). hen clculte V the volume inside the sphere nd inside the clinder 9 ( ) (see E.. Qu. ) V (volume of the entire sphere) V 6 9 +. he volume inside the sphere nd outside the clinder V V. he required volume V V V + V 8 9 + 8 9 6 9 +.

MAH 6 5.75.5.5.75.5 - -.5 -.5 -.5.5 E.., Qu. 7 f(,) + 5.5.5.5.75 E.., Qu..75.5.5 - -5 5 E.., Qu. 5 + f(,) ( + )( + ) 5-5 -.5.5.5.75...6 E.., Qu. f(,) +.8 -.5 -.5.5 -.5.5.5.75.5.5 - E.., Qu. 6 pge, Qu. 5