Changing the order of a triple integral

[JD_PM]

Homework Statement

$$\int_{-23/4}^4\int_0^{4-y}\int_0^{\sqrt{4y+23}} f(x,y,z) dxdzdy$$

Change the order of the integral to

$$\iiint f(x,y,z) \, \mathrm{dydzdx}$$What I have done

It is just about:

From $x=0$ to $x=\sqrt{4y+23}$

From $z=0$ to $z=4-y$

From $y=\frac{x^2-23}{4}$ to $y=4$

So

$$\int_0^{\sqrt{23}}\int_0^4\int_{\frac{x^2-23}{4}}^{4} f(x,y,z) dydzdx$$

But this is incorrect. What am I missing?

Thanks

 

[Mark44]

Homework Statement

$$\int_{-23/4}^4\int_0^{4-y}\int_0^{\sqrt{4y+23}} f(x,y,z) dxdzdy$$

Change the order of the integral to

$$\iiint f(x,y,z) \, \mathrm{dydzdx}$$What I have done

It is just about:

From $x=0$ to $x=\sqrt{4y+23}$

From $z=0$ to $z=4-y$

From $y=\frac{x^2-23}{4}$ to $y=4$

So

$$\int_0^{\sqrt{23}}\int_0^4\int_{\frac{x^2-23}{4}}^{4} f(x,y,z) dydzdx$$

But this is incorrect. What am I missing?

Thanks

Are you missing a sketch of the region of integration?
This region is fairly complicated, being the intersection of half of a parabolic solid whose axis is perpendicular to the x-y plane and a plane that makes an angle with the x-y plane.

 

[BvU]

What am I missing?

The integral over $dx$ yields a function of $y$
Try it with $f(x,y,z) = 1$

 

[JD_PM]

The integral over $dx$ yields a function of $y$
Try it with $f(x,y,z) = 1$

I would say $f(x,y,z) = 1$ is not relevant for the problem; I am just interested in changing the extremes and not in solving the integral.

 

[Mark44]

So

$$\int_0^{\sqrt{23}}\int_0^4\int_{\frac{x^2-23}{4}}^{4} f(x,y,z) dydzdx$$

But this is incorrect. What am I missing?

I ask again, did you draw a sketch of the region of integration?

The limits of integration for z (the middle integration) are wrong. The top of the solid is not horizontal, so z is not uniformly equal to 4.

 

[SammyS]

I would say $f(x,y,z) = 1$ is not relevant for the problem; I am just interested in changing the extremes and not in solving the integral.

Think about it.

Using $\ f(x,y,z) = 1\,,\$ as suggested by @BvU, would give a check as to whether the limits of integration you obtained give the same result as the original limits.

 

[JD_PM]

I ask again, did you draw a sketch of the region of integration?

The limits of integration for z (the middle integration) are wrong. The top of the solid is not horizontal, so z is not uniformly equal to 4.

The sketch is something like this (slice perpendicular to x axis, as asked). This is what I visualise now

x axis extremes: From $x=0$ to $x=\sqrt{23}$

z axis extremes: From $z=0$ to $z=4-y$

y axis extremes: From $y=\frac{x^2-23}{4}$ to $z=4-z$

But there is obviously something wrong with either the y or z extremes...

 

[Mark44]

The new order of integration is y, z, and finally, x, so it's best to list the limits on those three variables in that order. I have quoted what you wrote, but have rearranged the order in what I've quoted.

View attachment 239631
The sketch is something like this (slice perpendicular to x axis, as asked). This is what I visualise now

Nice drawing!

y axis extremes: From $y=\frac{x^2-23}{4}$ to $z=4-z$

No. Look at your drawing. The y values at the ends of the triangle base run from a point on the parabola to the line y = 4 in the x-y plane. The volume element is a cube of size dy*dz*dx. If you visualize a stack of these cubes, the stack will taper down to 0 as the y values increase to 4.

z axis extremes: From $z=0$ to $z=4-y$

Yes.

x axis extremes: From $x=0$ to $x=\sqrt{23}$

No, the x values run from 0 to the value at the point $(\sqrt{39}, 4)$. Do you see why?

 

[JD_PM]

Oh yeah so y-axis would be:

From $y=\frac{x^2-23}{4}$ to $y=4$

I do not see why you state those extremes for the x axis. I see how it goes from 0 to the parabola. Besides, you stated 3 values. Please explain your reasoning

 

[Mark44]

Oh yeah so y-axis would be:

From $y=\frac{x^2-23}{4}$ to $y=4$

Yes.

I do not see why you state those extremes for the x axis. I see how it goes from 0 to the parabola. Besides, you stated 3 values. Please explain your reasoning

Look at the triangle you drew in your sketch. As the triangle sweeps across from the y-axis, the y values range as you said above, and the x values go from 0 to the point on the parabola at which y = 4. IOW. to the point $(\sqrt{39}, 4)$.

 

[Ray Vickson]

View attachment 239631
The sketch is something like this (slice perpendicular to x axis, as asked). This is what I visualise now

x axis extremes: From $x=0$ to $x=\sqrt{23}$

z axis extremes: From $z=0$ to $z=4-y$

y axis extremes: From $y=\frac{x^2-23}{4}$ to $z=4-z$

But there is obviously something wrong with either the y or z extremes...

Your heavy black triangle is not perpendicular to the x-axis; it slants a bit towards the yz-plane.

 

[JD_PM]

Yes.
Look at the triangle you drew in your sketch. As the triangle sweeps across from the y-axis, the y values range as you said above, and the x values go from 0 to the point on the parabola at which y = 4. IOW. to the point $(\sqrt{39}, 4)$.

OK I've checked the answer:

$$\int_0^{\sqrt{39}}\int_0^{\frac{39-x^2}{4}}\int_{\frac{x^2-23}{4}}^{4-z} f(x,y,z) dydzdx$$

I understand the y-axis extremes.

Regarding z axis extremes; I get that its lower bound is $z=0$. Its upper bound is the plane $z=4-y$ ; why is $\frac{39-x^2}{4}$ the correct upper bound then?

Regarding x-axis extremes. I have the same trouble dealing with the upper bound of x; why is $\sqrt{39}$ ?

 

[Mark44]

Regarding z axis extremes; I get that its lower bound is $z=0$. Its upper bound is the plane $z=4-y$ ; why is $\frac{39-x^2}{4}$ the correct upper bound then?

The equation of the plane is z = 4 - y. The 3-D surface that the parabolic cylinder represents intersects the plane in a curve. Every point on this curve in the z = 4 - y plane lies above the parabola in the x-y plane, so these points satisfy both z = 4 - y and y = (x² - 23)/4. For the upper bound for the y integration, they have replaced y in the equation z = 4 - y, getting z = (39 - x²)/4.

Regarding x-axis extremes. I have the same trouble dealing with the upper bound of x; why is $\sqrt{39}$ ?

Again, look at the triangle you drew. These triangles start on the y-axis (with x = 0), and sweep out on the pos. x-axis to the point where the parabola crosses the line y = 4 in the x-y plane. What is the y-coordinate at that point?

 

[JD_PM]

You get $\sqrt{39}$, indeed. Thank you.