Question
find the volume bounded by the parabolic cylinder z=4-x^2 and the planes x=0, y=0, y=6 and z=0
I just want some help figuring out the limits of this question :
So for z, we have to integrate from z=0 to z=4-x^2 (am i right?)
W.r.t. y, we should integrate from y=0 y=6 (am i right?)
w.r.t. x , can you please explain?
Thanks!
I just want some help figuring out the limits of this question :
So for z, we have to integrate from z=0 to z=4-x^2 (am i right?)
W.r.t. y, we should integrate from y=0 y=6 (am i right?)
w.r.t. x , can you please explain?
Thanks!
Answers
The volume we want is in the 1st octant, since all the coordinate planes form part of the boundary. The cylinder is sliced off at y=6
So the volume is just 6 times the area under the curve z = 4-x^2
v = 6∫[0,2] 4-x^2 dx
If you insist on doing a triple integral, then you are correct as far as you go. The last step is just to integrate x from 0 to 2, since we want z=0 on the parabola.
So the volume is just 6 times the area under the curve z = 4-x^2
v = 6∫[0,2] 4-x^2 dx
If you insist on doing a triple integral, then you are correct as far as you go. The last step is just to integrate x from 0 to 2, since we want z=0 on the parabola.
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