Question

can u check it

Answer 1

aha check what

Answer 2

\[\int\limits_{0}^{a}\int\limits_{0}^{\sqrt{a^2-x^2}}\int\limits_{0}^{\sqrt{a^2-x^2-y^2}} xyzdxdydz\]

Answer 3

@Valerieeeee_8412
this one

Answer 4

are you trying to integrate f(x,y,z) = xyz over the sphere with radius 'a' ?
or just trying to find the volume of sphere ?

Answer 5

integrate
@hartnn

Answer 6

i don't think thats the original question...
the first limits are z = 0 to z = sqrt (a^2 -x^2- y^2)
which means you're first integrating w.r.t z
(then y, then x)
so ideally there should be
dzdydx
and NOT dxdydz as you wrote

Answer 7

okay so lets change it first

Answer 8

\[\int\limits_{0}^{a}\int\limits_{0}^{\sqrt{a^2-x^2}}\int\limits_{0}^{\sqrt{a^2-x^2-y^2}} xyz dzdydx\]

Answer 9

now that looks good :)

so you're first integrating w.r.t x,
then treat x and y as constants!

and \(\int z dz\)
is quite easy enough, right ?

Answer 10

yeah leave it

Answer 11

but what about the next integrate

Answer 12

first integrating w.r.t z ***

Answer 13

ok
what did u get after 1st integrationb ?

Answer 14

some thing like this ?
\((a^2-x^2-y^2)xy\)
with 1/2 taken out ?

Answer 15

yeah

Answer 16

its with root right

Answer 17

nopes

Answer 18

what is integration of z ?

Answer 19

y ??

Answer 20

z^2/2

Answer 21

\(\int z dz = z^2/2 +c\)

Answer 22

so, when you plug in the upper limit, it gets squared, right ?

Answer 23

okay then u apply the limits

Answer 24

correct

Answer 25

ohhhh yeah

Answer 26

now integrating w.r.t y, you'll treat x as constant

Answer 27

yeah

Answer 28

\((a^2-x^2-y^2)(xy) \)
first take 'x' out of integration

Answer 29

\(\int (a^2-x^2-y^2)y dy = \int (a^2-x^2)y -y^3 dy\)

Answer 30

that you can integrate ?

Answer 31

yeah

Answer 32

what do u get after integrating it ?

Answer 33

wait

Answer 34

\[(a^2-x^2)y^2/2 - y^4/4 \]

Answer 35

correct :)
now apply the limits

Answer 36

\( y^2 = a^2-x^2 \\
y^4 = (a^2-x^2)^2 \)

Answer 37

yeah

Answer 38

\[(a^2-x^2)^2 - 2(a^2-x^2)^2 /8\]

Answer 39

is it correct ?

Answer 40

why /8 ?
where did /2 of first term go ?

Answer 41

i multiply it with 4 to make in one part

Answer 42

so tell me how

Answer 43

\(\int x[ (a^2-x^2)^2/2- (a^2-x^2)^2 /4] dx\\ =(1/4)\int x(a^2-x^2)^2 dx \)

see if you get this

Answer 44

1/2 -1/4 =1/4

Answer 45

okay

Answer 46

than

Answer 47

that you can integrate ?
expand
(x^2-a^2)^2

Answer 48

okay surly thanks for that

Answer 49

you'll be able to finish it right ?
ask if any more doubts :)
welcome ^_^

Answer 50

@hartnn man

Answer 51

its should be looks like this
\[\int\limits x[ (a^2-x^2)^2/2- (a^2-x^2)^2 /4] dx\\ =(1/4)\int\limits x(a^2-x^2)^2 - (a^2-x^2)^2 dx\]

Answer 52

why is 'x' NOT multiplied to 2nd term ?

Answer 53

\(\int (a^2-x^2-y^2)xy dy =x[ \int (a^2-x^2)y -y^3 dy]\)

Answer 54

\[\int\limits\limits x[ (a^2-x^2)^2/2- (a^2-x^2)^2 /4] dx\\ =(1/4)\int\limits\limits x((a^2-x^2)^2 - (a^2-x^2)^2) dx\]

Answer 55

just check ur answer above

Answer 56

i am not sure where your algebraic doubt is...
maybe write it on paper ...
you should get
1/4 x (a^2-x^2)^2

Answer 57

okay whats the final answer should be ??

Answer 58

i am getting a^6/24

Answer 59

okay

Answer 60

\[1/8(\frac{ a^2 }{ 2 }(\frac{ a^2 }{ 2 }-\frac{ 2a^7 }{ 4 }+\frac{ a^2 }{ 10 }))\]

Answer 61

@hartnn

Answer 62

im getting this after i integrated x

Answer 63

thats incorrect
you cannot integrate
x and (a^2-x^2)^2 separately

Answer 64

x (a^2-x^2)^2 = x [a^4 +x^4-2a^2 x^2]

distribute

xa^4 +x^5 -...
and then integrate

Answer 65

ohhh i have to use uv method

Answer 66

okay okay this way i just know

Answer 67

ask if you still don't get a^6/24

Answer 68

man im not gritting that

Answer 69

getting **

Answer 70

@hartnn

Answer 71

x (a^2-x^2)^2 = x [a^4 +x^4-2a^2 x^2] = xa^4 + x^5 -2a^2 x^3

integrate this

Answer 72

\[x^2a^4/2+x^6/6-2a^2x^4/4\]

Answer 73

correct
plug in x=a the upper limit

Answer 74

for lower limit x=0, all terms =0

Answer 75

yeah then

Answer 76

im not getting seam answer just show me the steps

Answer 77

http://www.wolframalpha.com/input/?i=int_%7B0%7D%5E%7Ba%7Dint_%7B0%7D%5E%7Bsqrt%7Ba%5E2-x%5E2%7D%7Dint_%7B0%7D%5E%7Bsqrt%7Ba%5E2-x%5E2-y%5E2%7D%7D+x+y+z+dz+dy+dx

Answer 78

oh somewhere we missed 1/2

Answer 79

got it
in z^2/2
we missed this 1/2

Answer 80

yeah

Answer 81

but show me after ingratiated and put the limits of x

Answer 82

\(x^2a^4/2+x^6/6-2a^2x^4/4\)
is correct man,
and there is 1/8 out side

just plug in x=a in that
see what cancels out!

Answer 83

a^6/2 +a^6/6-a^6/2
=a^6/6

1/8(a^6/6) = a^6/48

Answer 84

yeah thanks a lot i got finally looolz

Answer 85

welcome ^_^