Question

Ok last question on my calc 3 homework and I can not seem to get the correct answer. Can anyone help me please?

Answer 1

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/6}\int\limits_{\sqrt{3}\sec \phi}^{4}3\rho^3\sin \phi d \phi d \rho d \theta\]

Answer 2

whenever i solve it i get 48.433 but apparently it is incorrect

Answer 3

sorry should be \[d \rho d \phi d\]

Answer 4

it might be easier to use abc instead of those greek things

Answer 5

the greeks makes it look like these are cylindrical coords

Answer 6

ok so \[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/6}\int\limits_{\sqrt(3)\sec(b)}^{4}3a^2\sin(b)dadbdc\]

Answer 7

err, sphericals ;)

Answer 8

haha ok well that is what this homework is covering spherical and cylindrical

Answer 9

so lets start with the innards\
\[\int_{\sqrt(3)\sec(b)}^{4}3a^2\sin(b)da\]

Answer 10

also, as a clean up, we can pull out the constants and associate them with their appropriate integrands

Answer 11

\[3\int_{0}^{2\pi}\left(\int_{0}^{\pi/6}\sin(b)\left(\int_{\sqrt(3)\sec(b)}^{4}a^2da\right)db\right)dc\]

Answer 12

alright looks good. the innermost one will actually cancle out with the 3 that you brought to the outside

Answer 13

so from there i got \[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/6}\sin(b)(64-(\sqrt(3))^3\sec^3(b)dbdc\]

Answer 14

\[3 \int_{0}^{2\pi}\left(\int_{0}^{\pi/6}\sin(b)\left(\int_{\sqrt(3)\sec(b)}^{4}a^2da\right)db\right)dc\]

\[3 \int_{0}^{2\pi}\left(\int_{0}^{\pi/6}\sin(b)\left(\frac134^3-\frac13\sqrt{3}sec(b)\right)db\right)dc\]

\[\int_{0}^{2\pi}\left(\int_{0}^{\pi/6}\sin(b)\left(64-3\sqrt{3}sec^3(b)\right)db\right)dc\]

\[\int_{0}^{2\pi}\left(\int_{0}^{\pi/6}64sin(b)-3\sqrt{3}tan(b)sec^2(b)~db\right)dc\]

\[\int_{0}^{2\pi}\left(-64\frac{\sqrt3}{2} - \frac32 \sqrt{3}~\frac43+64+\frac{3\sqrt{3}}2\right)dc\]

\[-\frac{8\sqrt3}{2} - 2\sqrt{3}+64+\frac{3\sqrt{3}}2~\int_{0}^{2\pi}~dc\]

something along those lines if i aint messed it up

Answer 15

ok um so that whole thing * 2pi?

Answer 16

i see i ran over a step :)
\[\int_{0}^{2\pi}\left(\int_{0}^{\pi/6}64sin(b)-3\sqrt{3}tan(b)sec^2(b)~db\right)dc\]
\[\int_{0}^{2\pi}\left(-64cos(30^o)-\frac32\sqrt{3}~sec^2(30^o)-(-64cos(0)-\frac32\sqrt{3}~sec^2(0))\right)dc\]

but yes

Answer 17

weird is saying it is wrong. every answer i have tried has been wrong so i guess i will email him about it. maybe the key is wrong

Answer 18

even wolframs triple integral calculator says it should be something i have tried already

Answer 19

http://www.wolframalpha.com/input/?i=integrate+%28from+0+to+2pi%29%28integrate%28from+0+to+pi%2F6%29%28integrate%28sqrt%283%29sec%28b%29to4%293a%5E2sin%28b%29da%29db%29dc

Answer 20

so what does the key say?

Answer 21

was the integral given to you? or did you try to define it with given information

Answer 22

it was given and idk what it says. its online and i only can only check if they are right or wrong. i emailed him earlier about it and he said "it looks like you have a sign switched"

Answer 23

abt 48.433 is what i keep getting from what youve posted.

i know some programs are picky if you dont input the answer in a specific format ..

Answer 24

are you able to take a screen shot and post it as a file attachement?

Answer 25

no thats not it because it converts it everthing to decimal numbers the grades the decimal and it only takes 5 places so that should be right

Answer 26

does it want an approximation? or the exact stuff with srts and pis?

Answer 27

either way is fine but i have tried both ways

Answer 28

\[(128-65\sqrt{3})*\pi\]

Answer 29

nope :/ its ok lol thank you for your help though

Answer 30

im pretty sure there is just an error with the key

Answer 31

GOOD LUCK ;)

Answer 32

opps, caps got locked