Question

(a) For the double integral pin2 (In 2)2-y I = ef+y* dx dy. i. Sketch the region of integration. ii. Show that I = (extu 2) (b) Using a triple integral, calculate the volume of the region in the first octant (x > 0, y > 0, z > 0), bounded by the two cylinders z2 + y2 = 4 and x² + y2 = 4.

Answer 1

Solution - We have given problem, Step-I Inz Ich22_y2 ez?ty ? dx dy (1) &x= (hom2) = ly inz & y= Что, 2t x=0 1 In 2 y = In 2 •693 x = Jen3 - yz yod y z X 1 2 (ii) We have given problem, y or Xo 4 we have to show, I = elin2) ² - 1) fa 2 sene - y² Now, I = pe ² + y² dx K 도 Z en 2 x² 4.4² 7+9 4 X = + 14 y² x + y2 e اما برای dy dn ..519 esc²ty²dx dy CS Scanned with CamScanner

Jln2²_y² te x²ty 0 0 As we know 2 j y dx dy Now convert the limit by Cylinder co-ordinate x² + y² = x=8coso N2 pln2 y Y arsino edz dr.do th u=82 la inde =2rdr = r dr ith all2 IT 2. NOW I du 1 Ana 2 2 & eddy.do I ett? du th > th a Hz eluz - ļ col du X2 17h 912 (eliz Th th I an 27 2 2 - -2 ¿ { + te du -1) du at hu biz I Iedz + Ins A { skus- i} Ane 2 T + Alna N { 2 ling CS Scanned with CamScanner