Question

use cylindrical or spherical coordinates to evaluate exactly the following

Answer 1

\[\int\limits_{0}^{a}\int\limits_{}^{(a^2-y^2)^1/2}\int\limits_{0}^{(a^2-x^2-y^2)} ~~~~x^2~~~~dzdxdy\]

Answer 2

\[\int\limits_{-2}^{2}\int\limits_{-(4-x^2)1/2}^{(4/x^2)^1/2}\int\limits_{(x^2+y^2)^1/2}^{(8-x^2-y^2)^1/2}\]

Answer 3

\[z^2~~~dzdydx\]

Answer 4

No offence, but your TeX needs work. You can write rational exponents by wrapping them in curly braces. Example: `a^b` gives \(a^b\), but `a^bc` gives \(a^bc\). To fix this, write `a^{bc}`, which gives what you want: \(a^{bc}\).

Answer 5

Also, what's the lower limit on the second integral of the first one you posted?

Answer 6

Oh Ok I will try to correct that. the limit is 0

Answer 7

No worries, I got it. Can you make sure I have it right?

First integral:
\[\int_0^a\int_0^{(a^2-y^2)^{1/2}}\int_0^{a^2-x^2-y^2}x^2\,\mathrm{d}z\,\mathrm{d}x\,\mathrm{d}y\]
Second integral:
\[\int_{-2}^2\int_{-(4-x^2)^{1/2}}^{(4-x^2)^{1/2}}\int_{(x^2+y^2)^{1/2}}^{(8-x^2-y^2)^{1/2}}z^2\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x\]

Answer 8

yes. it is.

Answer 9

I started working on the second one

Answer 10

I think we can use cylindrical coordinates?

Answer 11

Yeah, cylindrical would be a good idea. I'm trying to get a good visualization of the bounded region, just a minute...

Answer 12

And this is what I get for the second one. \[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{r}^{2} ~~~z^2~~~rdzdrd~\theta\]

Answer 13

We don't need to change z do we?

Answer 14

Not in cylindrical, no.

Answer 15

Here we are. Red indicates the solid region, orange and blue are the bounding surfaces.

Converting to cylindrical, you have
\[\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=z\end{cases}\implies \mathrm{d}V=\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z\]and the integral is
\[\int_0^{2\pi}\int_0^{\sqrt8}\int_{r}^{\sqrt{8-r^2}}z^2\,\mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\theta\]since we're going one revolution in the \(x\)-\(y\) plane; \(r\) starts at the origin and extends to the length of the sphere's radius; and \(z\) is bounded between the cone
\[\sqrt{x^2+y^2}=\sqrt{r^2}=r\]and the sphere
\[\sqrt{8-x^2-y^2}=\sqrt{8-r^2}\]

Answer 16

you forgot to include r

Answer 17

r dz dr d (theta )

Answer 18

How did you get \[\sqrt{8}\]

Answer 19

I am getting 2 for my bounds

Answer 20

\[y^2=4-x^2~~~~r^2\sin^2~\theta=4-r^2\cos^2\theta\]

Answer 21

solve that and you get 2

Answer 22

Oh I see, I'm mixing cylindrical with spherical. Sorry, I take by that comment.

Answer 23

Another correction:\[\int_0^{2\pi}\int_0^{\color{red}2}\int_{r}^{\sqrt{8-r^2}}rz^2\,\mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\theta\]

Answer 24

How does the sketch looks like