Triple Integral restricted to first octant.

In summary, the triple integral of function 14xz bounded between z=y^2 and z=8-2x^2-y^2 in the first octant can be solved using polar coordinates.
  • #1
PsychonautQQ
784
10

Homework Statement


Evaluate the triple integral of function 14xz bounded between z=y^2 and z=8-2x^2-y^2 in the first octant.


The Attempt at a Solution


So the first octant would mean the bottom parameter on all my integral will be zero since (x,y,z)>0. Then I set the equations equal to each other and got 4=x^2+y^2. It appears they intersect in a circle of radius two. I'm having a bit of trouble applying this knowledge to set up the limits of integration. Do I integrate with respect to Z first between 0 and x^2+y^2 and then evaluate dy and dx between zero and tw0?
 
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  • #2
Setting the two equations for [itex]z[/itex] equal is good. This gives you the projection of the curve, where the two surfaces intersect, to the [itex]xy[/itex] plane. You find that this is a circle around the origin. Thus, since in the first octand [itex]x,y[/itex] must run over the corresponding quarter-circle, and [itex]z[/itex] between the given boundaries.

Hint: It helps to draw the boundary surfaces to get a picture. Mathematica does a nice job, but I guess you can also do that with a free program like gnuplot.
 
  • #3
okay cool, so I did dzdxdy, dz was bounded by y^2 at the bottom and 8-2x^2-2y^2 at the top, and dy and dx were both from 0 to 2. Is that the correct way to set this up? If it is, screw me because I did the nasty integral by hand and rechecked my work a few times and got the same answer yet my online homework program is saying it's wrong
 
  • #4
PsychonautQQ said:
okay cool, so I did dzdxdy, dz was bounded by y^2 at the bottom and 8-2x^2-2y^2 at the top, and dy and dx were both from 0 to 2. Is that the correct way to set this up? If it is, screw me because I did the nasty integral by hand and rechecked my work a few times and got the same answer yet my online homework program is saying it's wrong

You have determined that the xy region is a circle of radius two. When you let both x and y go from 0 to 2, aren't you describing a square? You need to fix that.
 
  • #5
what if I make x go from 0 to (4-y^2)^(1/2) and then y go from 0 to 2? is that right?
 
  • #6
PsychonautQQ said:
what if I make x go from 0 to (4-y^2)^(1/2) and then y go from 0 to 2? is that right?

Yes. I'm assuming you have the correct integrand, but I don't know since you never showed your integral. You might find it easier to calculate using polar coordinates.
 
  • #7
is there a way to make wolfram alpha evaluate this X_X seems kind of nasty to do by hand... I'm integrating (14xy)dzdxdy
 
  • #8
in polar coordinates, would dz go from (rsin(theta))^2 to 8-2r^2? and d(theta) go from 0 to pi/2? and dr go from 0 to 8?
 

FAQ: Triple Integral restricted to first octant.

What is a triple integral restricted to the first octant?

A triple integral is a mathematical concept used to calculate the volume of a three-dimensional object. When it is restricted to the first octant, it means that the limits of integration are only in the positive x, y, and z directions, creating a volume in the first octant of a three-dimensional coordinate system.

What is the significance of restricting the triple integral to the first octant?

Restricting the triple integral to the first octant allows us to calculate the volume of a specific region in a three-dimensional space without including any negative values. This is useful in many scientific and engineering applications where we only need to consider positive values.

How is a triple integral restricted to the first octant solved?

A triple integral restricted to the first octant is solved by breaking down the volume into infinitesimal cubes and calculating the volume of each cube using the function that defines the shape of the region. These volumes are then added together to get the total volume of the region.

What are the limits of integration for a triple integral restricted to the first octant?

The limits of integration for a triple integral restricted to the first octant are typically defined by the boundaries of the region in the x, y, and z directions. For example, if the region is a sphere with a radius of 2 centered at the origin, the limits of integration would be x = 0 to 2, y = 0 to 2, and z = 0 to 2.

What are some real-life applications of a triple integral restricted to the first octant?

A triple integral restricted to the first octant has many real-life applications, such as calculating the volume of a solid object, determining mass and density of an object, and calculating the moment of inertia of an object. It is also used in fields such as physics, engineering, and economics to model and solve various problems involving three-dimensional spaces.

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