Question

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Answer 1

change to polar

Answer 2

x = rcos(theta)
y = rsin(theta)

Answer 3

solve z

Answer 4

step1 : convert to polar
step2 : scaling factor, dxdy = r dr dtheta
step3 : setup bounds

Answer 5

first express z in terms of r and theta

Answer 6

x^2/9 +y^2/9 +z^2/4 =1

z^2/4 = 1 - x^2/9 - y^2/9

z^2 = 4 [1 - x^2/9 - y^2/9]

z = 2 sqrt[1 - x^2/9 - y^2/9]

Answer 7

put x = rcos(theta)
y = rsin(theta)

Answer 8

z = 2 sqrt[1 - r^2/9]

z = 2/3 sqrt[9 - r^2]

Answer 9

we're done with step1.
see if you're at peace wid the stuff so far :)

Answer 10

so, the volume integral wud be :-

\(\large \mathbb{\int \int_R \frac{2}{3}\sqrt{1-r^2} ~ r dr d\theta} \)

Answer 11

we still need to setup the bounds ok

Answer 12

we're done wid first two steps. u seeing latex properly right ?

Answer 13

i made a typo, corrected below :-

\(\large \mathbb{\int \int_R \frac{2}{3}\sqrt{\color{red}{9}-r^2} ~ r dr d\theta} \)

Answer 14

how do we setup the bounds ?

Answer 15

u knw how the solid looks right ?

Answer 16

if its possible, can u draw it out ? :)

Answer 17

Excellent !!

Answer 18

nopes its more of a ellipsoidish.. :)

Answer 19

here is the volume integral :
http://www.wolframalpha.com/input/?i=int_0%5E%7B2pi%7D+int_0%5E%7B3%7D+2%2F3sqrt%289-r%5E2%29+r+dr+dtheta

Answer 20

looks good

Answer 21

it doesnt matter anyways, the shadow is simply a circle in xy plane.
we're interested in xy-plane oly

Answer 22

x^2/9 +y^2/9 +z^2/4 =1

for xy plane, put z = 0
u wil get a circle of radius 3.

Answer 23

do few more problems... these are easy ones... :)

Answer 24

u wil get hang of these in no time im sure. good luck !

Answer 25

Wowwww !!! thanks for the sweet testimonial XD