Triple integral, finding boundary

[jeffo]

Hi,
What would be the limits used in a triple integral that is bounded by the function 4x^2 + y^2 +z = 128, x=0, x=4, y=0,y=4?
Limits for x and y would be 0 to 4. Would the limits for z be 48 to (128-4x^2-y^2) or maybe 48 to 128?
Please let me know what limits to use, and the method of getting it.
Thank you.

 

[Dr.Peterson]

Hi,
What would be the limits used in a triple integral that is bounded by the function 4x^2 + y^2 +z = 128, x=0, x=4, y=0,y=4?
Limits for x and y would be 0 to 4. Would the limits for z be 48 to (128-4x^2-y^2) or maybe 48 to 128?
Please let me know what limits to use, and the method of getting it.
Thank you.

Where did 48 come from? Have you possibly omitted a condition?

The region as given is unbounded. If it had a "floor" of z=48, then you would integrate from 48 to 128-4x^2-y^2, assuming that is the innermost integral. (And thinking of "floor" and "ceiling" is my method.)

 

[jeffo]

Hi thanks for the reply, I got 48 by rearranging the 4x^2 + y^2 +z = 128, in terms of z: z = 128 - 4x^2 - y^2 and then subbing in x=4 and y=4 to give a lower limit of z. No conditions were omitted, this was all that was given :/. I agree with you, I plotted this and it seems there is no floor. is it possible the upper bound would be 128 and lower is 128-4x^2-y^2? Appreciate the help

 

[blamocur]

Where did 48 come from? Have you possibly omitted a condition?

The region as given is unbounded. If it had a "floor" of z=48, then you would integrate from 48 to 128-4x^2-y^2, assuming that is the innermost integral. (And thinking of "floor" and "ceiling" is my method.)

I am guessing 48 comes from [imath]128-4*4^2-4^2[/imath], i.e. value of [imath]z[/imath] at [imath]x=y=4[/imath].
Any chance that [imath]z[/imath] is squared too? I.e., an ellipsoid equation [imath]4x^2+y^2+z^2=128[/imath] ?

 

[jeffo]

I am guessing 48 comes from [imath]128-4*4^2-4^2[/imath], i.e. value of [imath]z[/imath] at [imath]x=y=4[/imath].
Any chance that [imath]z[/imath] is squared too? I.e., an ellipsoid equation [imath]4x^2+y^2+z^2=128[/imath] ?

Hi, thank you for the reply
nope the z is not squared, 3d plots suggest theres no lower bound i think which is strange. Perhaps there is something wrong with the question?

 

[BigBeachBanana]

Hi, thank you for the reply
nope the z is not squared, 3d plots suggest theres no lower bound i think which is strange. Perhaps there is something wrong with the question?

Can you post the ORIGINAL question?
If I were to guess the intent of the question...
[math]\int_{0}^{4} \int_{0}^{4}128-4x^2-y^2\, dydx=\int_{0}^{4} \int_{0}^{4}\int_{0}^{128-4x^2-y^2}1\,dzdydx[/math]

 

[jeffo]

Can you post the ORIGINAL question?
If I were to guess the intent of the question...
[math]\int_{0}^{4} \int_{0}^{4}128-4x^2-y^2\, dydx=\int_{0}^{4} \int_{0}^{4}\int_{0}^{128-4x^2-y^2}1\,dzdydx[/math]

This is the original question:
"A shape is bounded by the following elliptical function 4x^2 + y^2 +z = 128 and the planes x=0, x=4, y=0, y=4. Find the volume of the shape using triple integral. hint: Plot the geometry to help find the bounds"

 

[Dr.Peterson]

This is the original question:
"A shape is bounded by the following elliptical function 4x^2 + y^2 +z = 128 and the planes x=0, x=4, y=0, y=4. Find the volume of the shape using triple integral. hint: Plot the geometry to help find the bounds"

The equation there is an elliptical paraboloid. It's curious; if the z were meant to be squared, it would be an ellipsoid (but not a function), and the region would be bounded so you could solve the problem.

So it's hard to be sure where the error is, but there is an error in the problem.

 

[BigBeachBanana]

I agree with Dr.P. There needs to be another condition that bounds [imath]z[/imath] below. Otherwise, the volume is infinite.
The double integral I posted in #6 implicitly assumes that the distance between [imath]z=128-4x^2-y^2[/imath] and [imath]z=0[/imath]. Hence, the lower bound [imath]z=0[/imath] in the triple integral.

[math]\int_{0}^{4} \int_{0}^{4}128-4x^2-y^2\, dydx=\int_{0}^{4} \int_{0}^{4}\int_{0}^{128-4x^2-y^2}1\,dzdydx[/math]

However, that's still an assumption that was not given explicitly by the question. Therefore, I would say the question is faulty.