Question

calc 3 help!!!

use a triple integral to find the volume of the solid enclosed by the sphere x^2+y^2+z^2=2a^2 and the paraboloid az = x^2+y^2 (a>0)

Answer 1

first find the point of intersection of surfaces
since x^2+y^2=az
put in sphere equation
az+z^2=2a^2
z^2+az-2a^2=0
solve above equation
so z=a
(from given situation the other root is neglected)
now using z=a in any of the above given equations of surfaces
x^2+y^2=a^2
using cylindrical coordinates
since x^2+y^2=r^2 so

Answer 2

i hope you can do this integral if not let me know :)

Answer 3

@mickifree12 can you do this ?

Answer 4

not really, looking at this problem and a spherical coordinates one

Answer 5

would I treat a as a constant o.0

Answer 6

yes it is just a radius

Answer 7

I'm stuck =[

Answer 8

where ? in the integral?
just integrate first with respect to z treat r as constant then integrate with respect to r (now treat it as variable) and then integrate with respect to thetha.

Answer 9

wait, I see what you did now, I was making your "plug az into az+z^2=2a^2" more complicated than it really was

Answer 10

I can't solve the z^2+az-2a^2=0 to the point where you got z=a

Answer 11

it is just quadratic equation
let me solve it

Answer 12

oohhh that's why I didn't see it, I rarely use the quadratic equation to the point where I don't even know it my memory xP

Answer 13

ok here we have

Answer 14

got it now?

Answer 15

yes, I wasn't sure if I solve for Z or A, but... where are you getting these pictures from o.0?

Answer 16

i am typing in software known as Mathtype used for typing Maths .then take picture and upload here :)

Answer 17

aaawwww, I thought you somehow had a pdf of a book and it was done out in the book =[[

Answer 18

i use it because it is easy to use and Maths is readable like textbook format!

Answer 19

why are you neglecting that z=-2a?

Answer 20

look at the picture
below

Answer 21

since it is bounded by sphere above and paraboloid below taking z=-2a we are going away from the reigion of our interest.

Answer 22

ok, so from x^2+y^2=r^2

Answer 23

would I make the x and y cos and sin?

Answer 24

do not need to do this will get complex .

Answer 25

we just needs bounds for r and thetha in polar integrals .

Answer 26

I thought when you convert to r and theta, you convert your x and y into cos and sin

that way when you have something like x^2+y^2, when you convert them, it becomes simply 1

Answer 27

I don't exactly know spherical/cylindrical coordinates that well since my teacher just went over them

Answer 28

read these articles in your textbook.these are important.

Answer 29

did u solve the integral?

Answer 30

No, still trying to figure out this cylindrical coordinates right now

Answer 31

How did you know that d theta was from 0-2pi?

Answer 32

because x^2+y^2=a^2
is circle of radius r and for circle thetha goes from 0 to 2pi.

Answer 33

aaahh, doing the integral right now, just messed up and did the first integration
with respect to r instead of z, reworking it now

Answer 34

for the final answer, would a still be present o.0?

Answer 35

yes it will be

Answer 36

I'm getting a super ugly integration right when I do my integration with respect to r

Answer 37

yes !

Answer 38

ok, just making sure I'm not completely off

Answer 39

is it....

Answer 40

[2a^2/3 - 1/3(2a^2-81)^(3/2) - (729/4)]2pi

Answer 41

no you made some mistake check it !:)

Answer 42

ggaarrrrrrr

Answer 43

after the first integration...

Answer 44

rsqrt(2a^2-r^2) - r^3/4?

Answer 45

after first integration u get rsqrt(2a^2-r^2) - r^3/a

Answer 46

yeah oops, accidentally wrote 4, instead of A

Answer 47

and then.... 2nd integration, before plugging in 0 and 9

Answer 48

-1/3 (2a^2-r^2)^(3/2) - r^4/4a

Answer 49

yes

Answer 50

after plugging in

Answer 51

here's where it gets ugly

Answer 52

-1/3 (2a^2-81)^(3/2) - 6561/36) - (-1/3 (2a^2))

Answer 53

after pluging in you should have
1/12(8sqrt(2)-7)a^3

Answer 54

ok let me do it

Answer 55

I'm currently redoing it, from that point

Answer 56

I'm still getting the same answer, unless you somehow did some factoring o.0

Answer 57

wait

Answer 58

@mickifree12

Answer 59

for dr
is it 0-a o.0

Answer 60

this whole time I thought it said 0 to 9 -.-

Answer 61

yes

Answer 62

oh jeez ahahahah

Answer 63

i hope you can do it now :)
best of luck for rest of ur studies.!