Changing Order of Integration in Triple Integrals

[boneill3]

Homework Statement

Given

\int{0}{3} \int{0}{sqrt{(9-z^2)}} \int{0}{sqrt{(9-y^2-z^2)}} dx dy dz

express the integral as an equivalent in dz dy dx

Homework Equations

The Attempt at a Solution

From the origial limits z goes from 0 to when x^2 + y^2 <= 9

For dz I have

\int{0}{ sqrt(-x^2-y^2+9)} dz

Is this right so far?

 

[HallsofIvy]

Homework Statement

Given

\int{0}{3} \int{0}{sqrt{(9-z^2)}} \int{0}{sqrt{(9-y^2-z^2)}} dx dy dz

express the integral as an equivalent in dz dy dx

Homework Equations

The Attempt at a Solution

From the origial limits z goes from 0 to when x^2 + y^2 <= 9

For dz I have

\int{0}{ sqrt(-x^2-y^2+9)} dz

Is this right so far?

In the original integral z goes from 0 to 3, for each z, y from 0 to $\sqrt{9-z^2}$, and, for each y and z, x goes from 0 to $\sqrt{9- x^2- y^2}$.

That last is, of course, the upper hemisphere of $x^2+ y^2+ z^2= 9$ while the y integral is over the upper semi-circle of $y^2+ z^2= 9$. Finally the z integral over a radius of that circle and sphere. This is the first "octant" of the sphere of radius 3. Yes, If you change the order of integration so you are integrating with respect to z first, z goes from 0 to $\sqrt{9- x^2+ y^2}$.

Because of symmetry, this is a very easy problem!

 

[boneill3]

For dx

Do you leave the limit as \int{0}{sqrt{(9-y^2-z^2)}}dx or do you change it to

\int{0}{3} dx


\int{0}{3} \int{0}{sqrt{(9-z^2)}} \int{0}{ sqrt(9-x^2-y^2)} dz dy dx


regards

 

[HallsofIvy]

For dx

Do you leave the limit as \int{0}{sqrt{(9-y^2-z^2)}}dx or do you change it to

\int{0}{3} dx


\int{0}{3} \int{0}{sqrt{(9-z^2)}} \int{0}{ sqrt(9-x^2-y^2)} dz dy dx


regards

Since the x integral is the "outer" integral here, its limits of integration must be numbers, not functions of y and z! Yes, the limits are 0 and 3.

My point about the symmetry is that since this is an octant of a sphere, all choices of order are the same: except for "permuting" x, y, and z, those are exactly the limits of integration you had originally.

 

[boneill3]

Thanks