Triple Integration: (Someone to check my approach please)

Find the volume of the solid bounded by the cylinder x^2+y^2=4 and the planes y+z=4 and z=0 (i.e. xy plane)

My approach is below..

Question

Triple Integration:  (Someone to check my approach please)

Find the volume of the solid bounded by the cylinder x^2+y^2=4 and the planes y+z=4 and z=0 (i.e. xy plane)

My approach is below..

anonymous · Answer

$$V=\int\limits_{0}^{6}\int\limits_{-2}^{2}\int\limits_{-2}^{2} 1 dxdydz$$

anonymous · Answer

And I ended up with 93 units cubed

anonymous · Answer

it's been a bit since i've done these.. but I'm pretty sure you will need to use cylindrical coordinates somewhere there. i.e. you're going to need r and theta

anonymous · Answer

Yes i also thought so, but it confuses me so much so I tried it in cartesian form...

anonymous · Answer

Yeah I know what you mean I'm currently fighting with my brain to bring back this stuff aha.
well you know x^2+y^2=4 right?
that also means r^2=4 or r = 2
so 0<=r<=2 
and its a full circle so 
0<=theta<=2pi

anonymous · Answer

yep thats great thanks...except what difference will the other planes make on these limits...thats why i was cheeky and tried to keep it in cartesian....

anonymous · Answer

then maybe.... z =0 and z=4-y
so maybe you can use those as limits?

anonymous · Answer

and y would be r*sin(theta)

anonymous · Answer

Oh yes excellent point given that polar would require dr dtheta dz thanks!

anonymous · Answer

Ill give that a go and compare with my cartesian approach!

anonymous · Answer

alright good luck!

anonymous · Answer

Thank you!

anonymous · Answer

hmm do you think i should do it in the order dzdthetadr given that otherwise ill end up with a theta in the final answer?

anonymous · Answer

theta should be your last bound, z should be your first i think.

anonymous · Answer

ao dzdrdtheta?

anonymous · Answer

yeah. and remember when you're doing polar coordinates for that circle  its not just dr, you add another r so r*dr
dz*r*dr*dtheta

anonymous · Answer

excellent point thanks!

anonymous · Answer

okay- I got 200/3 pi hopefully I havent made any mistakes!

anonymous · Answer

as in 200pi/3

anonymous · Answer

do you have a way to check your answer?

anonymous · Answer

um..i wonder if wolfram alpha can do it?

anonymous · Answer

I wouldnt know what to type in thought...

anonymous · Answer

though*

anonymous · Answer

aha oh probably, yeah that's the battle with wolfram.

anonymous · Answer

I'll have a go...

anonymous · Answer

was this the integral you drew?

anonymous · Answer

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