Question

Z=0 (0,2), (1, 1), (2,2), and (1,3). 3) Find the volume of the solid that is lies between the cilinder x²+ x² + y² =4 and planes and ·Z = 3 4) Use the transformation u = x-y, V= x+ to evaluate SS dA G x+y where G is the square with vertices x-y

Answer 1

1 Answer: (3) Given that the solid bounded by cylinder x + y = 4 and planes 7=0 and y+z=3 Since y+z=3, we have 7= 3-4 'Z' varies from 'o' to '3-y' Since xv + y = 4, we have x = = 14-yv 'x varies from -√4-yr to 54-yr Here 'y varies from -2 to 2 x+y = 4 and if x=0, y = 4 - y = 12 Note that the volume of the solid cylinder bounded by the furfaces z=f(x,y) >0 and above a region & in xy-plane Now J4-yr (3-4) dx √4-yr y=-2 ソニー is given by SS zdady (or) SS feruldady Sszdedy = dy (223 ('64[ s« (3-4) [ JA -ym - E14-yr)] dy √4-yv X dy -J4-yv y=-2 2 y=-2 2 (3-4) J4-yndy y=-2

2 (3 14-yr) dy - q consider, I, = 3/2z-yudy 2 [I, - Ig] say Jax-yudy = 1 y Jac yr + Į a Sint (2)+c 2 q -=[# 3133 *****) -34 2 24x +* **s*et+) - 4(a ) ] - 1 / 2 Sin' (-2) 3[0+ 2. S'(1) - 0 - 8. Sinico)] 3[2. - 269) = (: Sin (-0) =- sino Sin' (1) = III 3.21 I = 67

and consider Iga ( ² y Jayrdy 3. let 4-glut differentiate w. v. to 't'm both sides =) D- 24 du = ydy - et Q 4-Ealat t=0 if y=-2, and il y = 2 4-2r=t t=0 It means that Ig=0 another words, s fl'dx= 0 it f(x) is an add function -a clearly the integrand in Ig is an odd function frufe - 9,5 s y J4-yddy = 0 - Ig Now SS z dady = 2 [I, - Ig 2[(T-0] = 12T .: The volume of the folid that lies b/w cylinder and planes - 12T

A (4) let the given transformation be u=x-y, v=x+y and that G is the given with vertices (0,2), square (1), (2, 2) and (13) Now, we substitute x-y=U and x+y= v to get (u, v) Then the new vertices of the square will be (u, v) (x-y, x+y) = (0-2,0+2) -) uw) = 62,2) at (0,2) at (11) (u, v) = (-1, I+I) = (0,2) at (2, 2), (u, v) = (2-2, 8+2) = (0,4) and at (13) (u, v) =(1-3, 1+3) = (-2,4) Observe that -aluco, 2<V 24 V (0,4) 4 3 9 (+2,2) 0,2) > u 1 2 3 4 we have x-y= u and x+y=re, Solve for x andy, weget x= utro علما and y= vu 2 2

5 The Jacobian of 'x and 'y' w. rto l'and'r, given by dx,y) IX dy J- Əx dx Ju do Jy Jx Oy Ju du ne al 2(4,7) he ди для =) J- Q = 6)-(+)61) -++-- x-Y dA Now s started SS dA G D where is the new domain containe (u, v) 4 u do du or u=-2 v=2 4 0 udu 1 do u=-2 V=2 4 는 2 0- 4 (love -slow) = (-2) Ina = ln (4) - dA = ln(6) iadna sta X-Y Х+У