Question

Find the volume under the given surface z=f(x, y) and above the rectangle with the given boundaries. ху 2 1sx52, 1 sys4 - (x² + y2) 2 Evaluate the integral with the given bounds Sl car mes3 duay = 0

Answer 1

solution - 157152, 15y34 __Given Boundries کے حج for the ada Take the content out I t. * mense die Apply 4 = y 2x2 & dy axda 7 dx = L dy

Simplity Adjust integral boundrill X=1 4 = y2H X24 % U = y2 +16 186 , du A6 I dy 72+26 Apply the power Rule frack xat att of 2+1 784 716 at 2

Simplity 2+16 --02- compute boundries जाजजामा-या ____426 (4)) - 40 (Fla)) lims -4) - - lim +2+ 2+16_241 2 S मो )

Now se prepotepung) % fake constant out в J, Take у 2+1)/7246) partial praction ч create partial fraction templete using the denominater — (91H) 2 +24) = ay+ay + 239-22 18 м) ?alb) 2 + 6 Muusibly Equation by sens. gl y2+)*26) - (ау ғал) (узы) (азые) (22H) Ca2+6) , **) Пазу +42) 72-1)(78) - yoto]

Simplity: y al aigtas)/y2 +16) +(234 +a)/29241) erband y = y3 +1 62 ft Qoy?+ 1620 +226 +237 +9242 +22 Group elements according to toy = 53/91 tas) + y2/aota2 +8 1694 powers +(169ota2) Equate the coefficient of . ? similar terms on both sides to create a list of & est of Equations T 1621taz to 7 1621 ta3 = 0 aota2 = 0 alta3 =0 - 23 = - 1 / 1 9 1 = 1/5 22 to 20²0 plug the solutions to the bestial tractions basameteal to obtain the sout

Lg to leto 15 g 2716 ge I SLy21) 15142716) Apply My Jane Rull I flad tg (1) da= f fla) da I f glodda fi signe bals) =(nie 1 (11/20) -6.07 = iş 6/8) – 1/2) = 4, { [0|20)-1914) 2 L 30 Alter Simplitilation 1015 - 10/21 - 10120)+10 (17) = 0.18844