Question

Use a triple integral to find the volume of the solid shown in the figure.

Answer 1

\[z = 64 −x ^{2} − y ^{2}\]

Answer 2

Any hints or help is welcome.

Answer 3

are*

Answer 4

I saw this problem , just don't have time to review. if you just want a hint, I can!! change to polar system to \(\rho ~~\phi ~~\theta\) don't forget jacobian number

Answer 5

hey, camoJAX. I remember when I took cal3, you were at the same level with me. Why are you asking the same thing which we studied 1 year ago?

Answer 6

I wasn't doing Cal3 a year ago, DiffEq yea

Answer 7

\[

V=\iiint dV
\]

Answer 8

\[
0<z<64-x^2-y^2
\]

Answer 9

the problem I have is I can never understand where the integrals should start or end

Answer 10

Well your boundaries should look like \[
f_1(x,y)<z<f_2(x,y)\\
g_1(x)<y<g_2(x)\\
a<x<b
\]

Answer 11

In this case, because the 3D solid is z simple. It's 2d projection onto xy is y simple

Answer 12

So the order is z, y, x. You could also do z, x, y

Answer 13

hard to tell what the radius of the circle on the bottom is

Answer 14

an educated guess would be 8. if that's the case:

\[\int\limits_{-8}^{8}\int\limits_{-\sqrt{64 - x^2}}^{\sqrt{64 - x^2}}\int\limits_{0}^{64 - x^2 - y^2}dzdydx\]

and you start from the inside out, so dz, dy then dx

Answer 15

that's confusing me also

Answer 16

makes sense. when y or x = 0, the other one would be 8. so it's 8

Answer 17

yea that does

Answer 18

at z = 0*

Answer 19

another way to look at this is with spherical coordinates. and it is actually simpler. however it really requires theory that is taught by a teacher or a textbook. if you have some knowledge of it i can set it up though

Answer 20

I honestly have no knowledge of it but I still would like to know if you don't mind

Answer 21

you will definitely learn it in class ^_^

Answer 22

haha okay thanks