Question

can you please answer both of these please

x = Su, y=6v, z =4w; S S S 22 dx dy dz, R where is the interior of the ellipsoid 384 5" x2 v2 22 64 3616 4811 6411 256 TT

Answer 1

x=84 Y = 6V Zaqw Jacobian is given by xley dylan 8 delar azlay dzlar dylav 2) 6 O show dylow dzlow O o 4 JE 8 X 6 X4 192 R: 22 64 + 22 36 36 64.42 64 f v2 + 1 w2al 4² + y2 + w2a 1 Ukring Espherical Coordinades har 4 = f sin(a) BLO) y = 8 sint) sin 10 f Copy u²+V2 at W2 = 82, audy dwa 8 sinds de do 18 = 1 of pat of Oz 25 Newe, 1 = SSS z 2 dx dy dza SS (A w) 2 /5) duardo 58(62) (192)) detro w R R +

I ne smyss w2d4dvd w R .24 =3072 82 6682 P. 8² sin 83 294 do I = (3042) S** 405* op 24sin (4) * J's+ d} +1 but Go bat -dta sin f&& क 21 I å 6072) ***644) No 1 = -3042 (22-9) (ime) ). 1 - 3072 x 25 T 5x 3 - 1024 너 x 27 x (-)-1) = 1024 x 41 4096 1

Prol 55 226, 22 y = 6 I = SS ry 4A = JS ay d y dy R R 6 6 al ST ny dy du af ndas gdy 2 3 2 17 G 3), (35724 36-25 136 4 2 32 2 11 X 32 - 1188 89

According to given details the answer of first question is

(4096/5)*pi

If any quadrant is given then the answer is

(256/5)*pi

2)

88