Question

Use triple integrals to find the volume of the solid E bounded by the parabolic cylinder z=1 - y2 over the square (-1, 1] x [-1, 1) in the xy-plane. Hint: Volume(E) = SSSE 1 DV Answer: 8 3 z=1 - 22 In each of the given orders, SET UP the integrals for a function f over the solid shown. If this can not be done using a single set of triple integrals, state NOT POSSIBLE. a) dx dy dz b) dz dy dx Answer: c) dy dx dz a) Not possible y y=1- x х b) 61155 (0,1,2) dzdyde CL si f(x, y, z) dydzdz 1-2 . 1-2

Answer 1

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... f: z = 1 - 2 + Input... 4 3

1360,0,1) ay cil dv (-1,-1, of are comidra small region of arbitrarily small volume dadydz vary the position of de along 2,4x aires respectively. OS ZSI-y? -leg علم (1,-1,0) 4 Y 1 Ein ginen by als as

Volume of E 1.42 dzdy da na yol z=0 Il LE - (S'S ser SCHL-u dla -yYdy da -7 -1 li / 2 2 dna 4 l da 3 So dx = 4/3 (1-1) -1 N zalog 3 Volume (E)=8 3 For the given region Ecsey) we convichen an arbitrary small rolume in E. say dr The dimensions of this volume is de, dy and dz Where عزة is du= dady dz We vary dv along X, Y, Z in different onder to obtain the limit of the sitegration. The limits are oftained by the boundary of E. a) dady dz Here are first vary the elementary volume along x arb we get osasiny Next we are vary to the Y. z. flane, along Y an's first. has po be a function of 2. Ele of his constant we won't get the required volume. b the elementary volume Parallal for this Y

but note that Zal-x : Z does not depend on y So it is not possible to find the integration by taking dady dz 3) dz dy da This case we first rary 65n31 the elementary voleme about Zaris, then about y and lastly about X asis. osysi-2 har dzdy da. So we have OSz51-22 . volime is a=0 Y=0 Z-0 This care c) dy da da we fiouf meine farallal to Yan's The's gives the range of Y to be osy <1-a. Next we move forallal to Nazis. the upper brit of a will dependon z limbke the est core) because I clepends on y. =Vr-2 taking the partive Granch so os neri-z of the paraboloid. Lastly we more parallal to Zaris. at top most part Z=10 so Osesi Hence the volume is dy dadz. so 1-=Z => a = (0)"=1 1-Z 1- SL 2-0 1-0 Y=0

The volume in So's corresponding i-n-23 dz dyda I-2 S'S. (1-2²) dyda So' (1-x)¢14 ») da (1-2x +27) (172) da So =So 1-2a+n+a-2ntu² da So' 1-x-x² + x3 da 11% -3% +3% =1-{ - 1/3 tt 3 11 5/12