Iterated integrals over region w

[jonroberts74]

Homework Statement

I am given

W = \{ (x,z,z)| \frac{1}{2} \le z \le 1; x^2 + y^2 +z^2 \le 1\}

they want the iterated integrals to be of the form

\iiint_W dzdydx

The Attempt at a Solution

so I know z=1/2 will give me the larger bound for x

x^2 + y^2 + (1/2)^2 =1 \rightarrow x^2 + y^2 = 3/4


x^2 + y^2 = 3/4 gives me - \sqrt{3/4} \le x \le \sqrt{3/4} when y=0

so y bounds are

- \sqrt{3/4-x^2} \le y \le \sqrt{3/4-x^2}

and my z bounds

\sqrt{1-y^2} + \frac{1}{2} \le z \le \sqrt{1-y^2}


so I get

\int_{- \sqrt{3/4}}^{\sqrt{3/4}}\int_{- \sqrt{3/4-x^2}}^{\sqrt{3/4-x^2}}\int_{\sqrt{1-y^2} + \frac{1}{2}}^{\sqrt{1-y^2}}f(x,y,z)dzdydx

 

[Orodruin]

I suggest a change to cylinder coordinates to make one of the integrals trivial.

The way you have written it, it is not clear what you intend to be the bounds for what integral.

 

[jonroberts74]

I suggest a change to cylinder coordinates to make one of the integrals trivial.

The way you have written it, it is not clear what you intend to be the bounds for what integral.

\int_{- \sqrt{3/4}}^{\sqrt{3/4}}\Bigg(\int_{- \sqrt{3/4-x^2}}^{\sqrt{3/4-x^2}}\Bigg(\int_{\sqrt{1-y^2} + \frac{1}{2}}^{\sqrt{1-y^2}}f(x,y,z)dz\Bigg)dy\Bigg)dx

maybe that will help

I'll try it out with cylindrical coordinates as well

 

[LCKurtz]

and my z bounds

\sqrt{1-y^2} + \frac{1}{2} \le z \le \sqrt{1-y^2}


so I get

\int_{- \sqrt{3/4}}^{\sqrt{3/4}}\int_{- \sqrt{3/4-x^2}}^{\sqrt{3/4-x^2}}\int_{\sqrt{1-y^2} + \frac{1}{2}}^{\sqrt{1-y^2}}f(x,y,z)dzdydx

Your z bounds are incorrect. z should go from the lower surface to the upper surface, which is a function of both x and y.

 

[Orodruin]

The integration bounds on z depend both on x and y. The lower bound on z is 1/2 regardless of x and y. The upper must be found for arbitrary x and y (within their bounds).

I really really suggest the cylinder coordinates. :)

 

[jonroberts74]

Your z bounds are incorrect. z should go from the lower surface to the upper surface, which is a function of both x and y.

hm, the way I did it was once I had x, I sliced with the x=0 plane so that I'd have a 2-D image then took the bounds of y and figured out how z changes in the plane from 1/2 to 1.

 

[jonroberts74]

I'll try cylindrical after my current problem

are the z bounds \frac{1}{2} \le z \le \sqrt{1-x^2-y^2}

if it relies on both that's what makes sense

 

[LCKurtz]

hm, the way I did it was once I had x, I sliced with the x=0 plane so that I'd have a 2-D image then took the bounds of y and figured out how z changes in the plane from 1/2 to 1.

Doing z first, you must go from the lower surface to the upper surface. Then you look in the xy plane for the "shadow" of the volume which, in this problem, is an ellipse. That's where you get the x and y limits. And what is $f(x,y,z)$?

 

[Orodruin]

Yes. If you fix x and y to arbitrary values within their common bounds, those are the resulting bounds on z.

 

[jonroberts74]

Doing z first, you must go from the lower surface to the upper surface. Then you look in the xy plane for the "shadow" of the volume which, in this problem, is an ellipse. That's where you get the x and y limits. And what is $f(x,y,z)$?

f(x,y,z) is just a place older, there's no given function. it just asks for it to be set up as iterated integrals.

in my course anytime we've been setting up integrals we use that as an arbitrary placeholder

 

[jonroberts74]

I think I converted that over correctly

$\displaystyle \int_{0}^{2\pi}\int_{0}^{1}\int_{1/2}^{1}r\cdot rdzdrd\theta = \frac{\pi}{3}$

not positive why there's an extra 'r' in the integral but that's what seems to make it correct.




I think this is is the integral I should have gotten

$\displaystyle \int_{0}^{2\pi}\int_{0}^{1}\int_{\frac{r}{2}}^{r}rdzdrd\theta = \frac{\pi}{3}$

 

[LCKurtz]

You're getting worse, not better.

I think I converted that over correctly

$\displaystyle \int_{0}^{2\pi}\int_{0}^{1}\int_{1/2}^{1}r\cdot rdzdrd\theta = \frac{\pi}{3}$

not positive why there's an extra 'r' in the integral but that's what seems to make it correct.

Look in your post #7. The upper limit for $z$ is not $1$. Nor is $r=1$ correct for its upper limit. And why the extra $r$?

I think this is is the integral I should have gotten

$\displaystyle \int_{0}^{2\pi}\int_{0}^{1}\int_{\frac{r}{2}}^{r}rdzdrd\theta = \frac{\pi}{3}$

And the upper limit for $z$ isn't $r$ either. And its lower limit isn't $r/2$. And the upper limit for $r$ isn't $1$. Perhaps you should just start over.

 

[jonroberts74]

You're getting worse, not better.
Look in your post #7. The upper limit for $z$ is not $1$. Nor is $r=1$ correct for its upper limit. And why the extra $r$?
And the upper limit for $z$ isn't $r$ either. And its lower limit isn't $r/2$. And the upper limit for $r$ isn't $1$. Perhaps you should just start over.

the extra r came simply from a heuristic approach, it seemed correct.

starting over:first thing converting over to cylindrical

$0 \le \theta \le 2\pi$ this one I see no issues with

I have $x^2+y^2+z^2 \le 1$

this is where I wasn't sure about converting, do I treat $z^2$ just as is or treat it as $z^2=x^2+y^2$

the first way of treating z

$r^2cos^2(\theta) + r^2sin^2(\theta) + z^2 =1$ but graphing this is not the same sphere, it
the second way

$r^2cos^2(\theta) + r^2sin^2(\theta) + (r^2cos^2(\theta) + r^2sin^2(\theta)) =1$ but in order for this to be the same sphere it has to be equal to 2 or I take out $(r^2cos^2(\theta) + r^2sin^2(\theta))$ and keep it set as equal to 1

some explanation on this would be appreciated immensely

now for the r and z bounds

$\frac{1}{2} \le z \le \sqrt{1-x^2-y^2}$ this does not work if put as $\frac{1}{2} \le z \le \sqrt{1-r^2cos^2(\theta)-r^2sin^2(\theta)}$

this can't be that difficult but I am missing something

EDIT: also doing it in rectangular I graphed my x,y bounds and those aren't any good either, only way I could get the proper bounds was to have each of x,y,z in terms of the other two.

 

[jonroberts74]

I am returning to this problem even though my semester is over. I believe spherical would be better than cylindrical.

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{1/2}^{1}\rho^2sin\phi d\rho d\phi d\theta$

 

[LCKurtz]

I am returning to this problem even though my semester is over. I believe spherical would be better than cylindrical.

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{1/2}^{1}\rho^2sin\phi d\rho d\phi d\theta$

Did you pass you course?

Your setup is still incorrect. $\rho$ is not constant on the lower surface $z=1$. And the upper limit for $\phi$ is not $\frac \pi 4$.

 

[jonroberts74]

Did you pass you course?

Your setup is still incorrect. $\rho$ is not constant on the lower surface $z=1$. And the upper limit for $\phi$ is not $\frac \pi 4$.

I managed to get in A in the course, I am still having difficulties with triple integrals though so I am continuing to work on them

z=1 is the upper surface isn't it? And I thought phi would go from 0 to $\frac{\pi}{4}$ because phi starts from the positive z axis.

I am looking at it in grapher. I have a unit sphere and two plane, z=1 and z=1/2, slicing through it.

$0 \le \theta \le 2\pi$ I find no issues with that

fpr $\phi$ it makes sense looking at it that it goes form 0 to $\pi/4$. I have a 2-D view of it graphed as well, like I sliced with the x=0 plane.

 

[LCKurtz]

I managed to get in A in the course, I am still having difficulties with triple integrals though so I am continuing to work on them

z=1 is the upper surface isn't it? And I thought phi would go from 0 to $\frac{\pi}{4}$ because phi starts from the positive z axis.

No. $z=1$ is the largest value of $z$ in the figure, but the upper surface is the sphere and the lower surface is the plane $z=1/2$. Draw a cross sectional picture of the sphere and plane to figure out the max value of $\phi$.

 

[jonroberts74]

isn't z=1 the largest value without the z=1 plane even in there? its a unit sphere so I don't even see why that plane is necessary.

looking at a cross sectional picture in the yz plane: I have a unit circle z^2+y^2 = 1 sliced by a line z=1/2

and phi starts at the positive z-axis and would stop where z=1/2 slices through so that's a movement of 0 to $\pi/3$

but $\rho$, $z=\rho cos(\phi)$ so $\rho = \frac{z}{cos(\phi)}$

 

[LCKurtz]

isn't z=1 the largest value without the z=1 plane even in there? its a unit sphere so I don't even see why that plane is necessary.

looking at a cross sectional picture in the yz plane: I have a unit circle z^2+y^2 = 1 sliced by a line z=1/2

and phi starts at the positive z-axis and would stop where z=1/2 slices through so that's a movement of 0 to $\pi/3$

Yes. That's better than just guessing that $\rho = \pi/4$ eh?

but $\rho$, $z=\rho cos(\phi)$ so $\rho = \frac{z}{cos(\phi)}$

But you want $\rho$ on the plane $z=1/2$. So $\rho$ there is ..?

 

[LCKurtz]

Homework Statement

I am given

W = \{ (x,y,z)| \frac{1}{2} \le z \le 1; x^2 + y^2 +z^2 \le 1\}

isn't z=1 the largest value without the z=1 plane even in there? its a unit sphere so I don't even see why that plane is necessary.

Look at your original statement. You want $z$ between $1/2$ and $1$, but also inside the sphere $x^2+y^2+z^2=1$. That region lies above the plane $z=1/2$ and below the upper hemisphere of $x^2+y^2+z^2=1$. Those are your lower and upper surfaces.

 

[jonroberts74]

Look at your original statement. You want $z$ between $1/2$ and $1$, but also inside the sphere $x^2+y^2+z^2=1$. That region lies above the plane $z=1/2$ and below the upper hemisphere of $x^2+y^2+z^2=1$. Those are your lower and upper surfaces.

for the upper limit of $\rho$,

$\rho = \frac{z}{cos\phi} = \frac{1}{cos(0)} = 1$

for the lower limit

$\rho = \frac{1/2}{cos\phi} = \frac{sec\phi}{2}$$\phi$ EDIT: $\rho = \frac{1/2}{cos\phi} = \frac{sec\phi}{2}$

so $\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{\frac{sec\phi}{2}}^{1}\rho^2 sin\phi d\rho d\phi d\theta$

 

[LCKurtz]

for the upper limit of $\rho$,

$\rho = \frac{z}{cos\phi} = \frac{1}{cos(0)} = 1$

No, you aren't getting it. Right answer (just luck) but definitely wrong reason. The upper limit is $\rho$ on the sphere. What is the equation of the sphere in spherical coordinates?

for the lower limit

$\rho = \frac{1/2}{cos\phi} = \frac{sec\phi}{2}$[STRIKE]$\phi$[/STRIKE]

I assume that extra $\phi$ is a typo.

so $\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{\frac{sec\phi}{2}}^{1}\rho^2\color{red}{\sin\phi} d\rho d\phi d\theta$

I corrected your dV element for spherical coordinates. As I said above, the limits are now correct but you haven't correctly explained why the upper limit for $\rho$ is $1$.

Once you correctly explain that, since you got an A in the calc III course, my challenge to you is to work out this integral. Then set it up correctly in cylindrical coordinates and work it also to get the same answer. Are you up to the challenge?

 

[jonroberts74]

No, you aren't getting it. Right answer (just luck) but definitely wrong reason. The upper limit is $\rho$ on the sphere. What is the equation of the sphere in spherical coordinates?
I assume that extra $\phi$ is a typo.
I corrected your dV element for spherical coordinates. As I said above, the limits are now correct but you haven't correctly explained why the upper limit for $\rho$ is $1$.

Once you correctly explain that, since you got an A in the calc III course, my challenge to you is to work out this integral. Then set it up correctly in cylindrical coordinates and work it also to get the same answer. Are you up to the challenge?

yes the $\phi$ was a typo I had written something out but deleted it all except for that it appears. and I was going back to add in the $sin\phi$ I missed

the equation in spherical is $\rho^2 = 1$ so $\rho = 1$ But I don't see what was wrong with the way I got 1 as the upper limit I using $\rho = \frac{z}{cos\phi}$ I didn't just guess it was 1, and I worked out what the lower limit was.

 

[Orodruin]

Just to say something with regards to spherical coordinates vs cylinder coordinates: as you noticed, the surface of the sphere was a simple coordinate surface in spherical. This will no longer be true in cylinder coordinates, so this will complicate life a bit. On the other hand, something else might be a bit easier.

 

[jonroberts74]

$\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{\frac{sec\phi}{2}}^{1}\rho^2 sin\phi d\rho d\phi d\theta$

$\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \frac{\rho^3}{3} sin\phi \Bigg|_{sec\phi/2}^{1} d\phi d\theta$$\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} (\frac{sin\phi}{3} - \frac{1}{24}tan\phi sec^2 \phi) d\phi d\theta$$\displaystyle \int_{0}^{2\pi} (\frac{- cos\phi}{3} - \frac{1}{48} sec^2 \phi)\Bigg|_{0}^{\pi/3}d\theta$

$\displaystyle \int_{0}^{2\pi} \frac{5}{48} d\theta$

$=\frac{5\pi}{24}$

 

[LCKurtz]

But I don't see what was wrong with the way I got 1 as the upper limit I using $\rho = \frac{z}{cos\phi}$?

Because that equation is true for any point in space when expressed in spherical coordinates. How do you get $\rho$ is constant on the sphere from that?

 

[LCKurtz]

$\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{\frac{sec\phi}{2}}^{1}\rho^2 sin\phi d\rho d\phi d\theta$

...

$=\frac{5\pi}{24}$

Good. Now can we see it in cylindrical coordinates?

 

[jonroberts74]

Because that equation is true for any point in space when expressed in spherical coordinates. How do you get $\rho$ is constant on the sphere from that?

I can see that, but I know the plane z=1 intersects the unit sphere at z=1 and phi is cos(0) at the z-axis so I used that, I did use \rho^2 = 1 on my scratch sheet I just didn't include it in my post. I'll be more careful in the future. I have to take care of some errands but I'll come back and do cylindrical.

 

[jonroberts74]

It's been a busy day but

I worked it out on scratch paper

$\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{3}}{2}} \int_{\frac{1}{2}}^{\sqrt{1-r^2}} rdzdrd\theta$

I got the same answer of $\frac{5\pi}{24}$

I want to try is solve it as if it set as $W=\{(x,y,z)|\frac{1}{2} \le x \le 1; x^2+y^2+z^2=1\}$ using cylindrical and not treating the x-axis as the z-axis because then the above answer would be the answer to it. I assume it is possible

 

[LCKurtz]

It's been a busy day but

I worked it out on scratch paper

$\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{3}}{2}} \int_{\frac{1}{2}}^{\sqrt{1-r^2}} rdzdrd\theta$

I got the same answer of $\frac{5\pi}{24}$

Good.

I want to try is solve it as if it set as $W=\{(x,y,z)|\frac{1}{2} \le x \le 1; x^2+y^2+z^2\color{red}{=}1\}$ using cylindrical and not treating the x-axis as the z-axis because then the above answer would be the answer to it. I assume it is possible

Yes. You should be able to do that and get the same answer. You want $\le 1$ for the sphere. Your $r,\theta$ limits will be similar to what you did for $\rho,\phi$ in spherical. Post what you get if you run into trouble.