Question

Find the volume of the region inside z^2=x^2+y^2 between z=1 and z=9. Is the z limits 1 < z < 9... and not really sure on the others. Seems simple but I'm confused.

Answer 1

you have a sphere with a section between 1 and 9 is what it looks like to me

Answer 2

its a cone aint it

Answer 3

so 1<z<9, 1<y<3 and -rad(z^2-y^2)<x<rad(z^2-y^2) I assume that is wrong...

Answer 4

dunno yet, we will determine that after I get a good bead on it

Answer 5

spose x=0; the z^2 = y^2
z = +- sqrt(y^2)
= +- y

Answer 6

and it is a cone

Answer 7

same calcs for the y=0 parts and we get a cone about the z axis

Answer 8

is it easiest to use polar coordinates or cartesian?

Answer 9

for this, either or ...

since the radius of each circle is defined by x or y; we can simply say:
\[A=\pi [f(x)]^2\]

or f(y) ^2 but we have to pick one of them

Answer 10

cylindrical coords is an option too; but they all amount to the same thing i believe

Answer 11

the radius moves from x=0 to x=sqrt(z)

Answer 12

\[\int_{1}^{9}\int_{0}^{\sqrt{z}}\ 2\pi\ x dx.dz\]
should do it

Answer 13

I understand the limits. But why do you have 2*pi*x? Sorry I'm new to this.

Answer 14

quite frankly, i had to adjust my original thoughts :)

essentially this adds up all the circumferences from the center out; the sum of all the circumferences add up to the area; then we do for all the points along z from 1 to 9

Answer 15

add up all the circumferences to get area of 1 circle

Answer 16

the area of each circle is: \(z\ \pi \)

int z = z^2/2 from 1 to 9 = 9^2/2 - 1^2/2

Answer 17

so 40?

Answer 18

40 pi :)

Answer 19

I get the way you figured it out but how would you formally set it up as a double integral.

Answer 20

one way is the way I did it; im sure there are other ways of course

Answer 21

we could define a triple as the radius, the angle, and the z axis

Answer 22

\[\int_{1}^{9}\int_{0}^{2pi}\int_{0}^{\sqrt{z}} dr.d\theta .dz\]

Answer 23

That makes perfect sense. Thanks!! It should be "r*dr*d(theta)*d(z)" though right?

Answer 24

i believe it starts with the function r=1

Answer 25

Could you explain that? I don't understand what you mean it stats with the function r=1.

Answer 26

lets try it

: dr = r, sqrt(z) - 0 = sqrt(z)

: sqrt(z) dt = sqrt(z) t ; 2pi sqrt(z) - 0

: 2pi sqrt(z) dz = 4pi/3 z^(3/2)

hmmm, i could be wrong :) but yeah, it works out something along these lines


integration is just adding up alot of points, so as long as you know what your adding up and how it moving; youre good to go

Answer 27

the circumference of a circle = 2pi r , say we want a radius of 5
\[\int_{0}^{2pi}\int_{0}^{5}dr.dt=\int_{0}^{2pi}r;[0,5]\ dt\to\ \int_{0}^{2pi}5dt\]
\[\int_{0}^{2pi}5dt\to\ 5t;[0,2pi] = 10pi\]

ok, so that route gets us to circumferece in the end

Answer 28

oh well; id stick with a double of 2pi r; r=0 to sqrt(z) and z goes from 1 to 9

Answer 29

Thank you, sir.

Answer 30

yw