Calculate a Triple Integral using the cylindrical coordinate system

[Gabry89]

I don't understand what's wrong with it:

 

[tiny-tim]

Hi Gabry89! Happy new year!

In your condition 0 ≤ y ≤ 1 - x - z,

you need to deal separately with the case x + z > 1.

 

[Gabry89]

thanks, Happy new year too ;)

I thought about that, here's what I've done:

 

[tiny-tim]

no, that doesn't work …

you're integrating ρ from 1/(cosφ + sinφ) to 1 …

which doesn't work for cosφ + sinφ < 1

as i said, deal separately with the case x + z > 1.

 

[Gabry89]

^i think I've understood what you mean but i just don't realize where's the case x+z > 1, i just see x + z <= 1

 

[Char. Limit]

^i think I've understood what you mean but i just don't realize where's the case x+z > 1, i just see x + z <= 1

I don't know about you, but I see [tex]x^2 + z^2 \leq 1[/tex].

 

[Gabry89]

^yes, that's not what i mean, i mean that

0<= y <= 1-x-z then x+z <= 1 not x+z > 1 as the case that tiny-tim was talkin' about

 

[tiny-tim]

… i just don't realize where's the case x+z > 1, i just see x + z <= 1

Exactly … you've dealt with x + z > 1 in the same way as x + z ≤ 1, getting a negative integral (ie with the "upper" limit lower than the "lower" limit) for the second one.

You need to write the integrals separately, so that you can put the second one equal to zero.

 

[Char. Limit]

^yes, that's not what i mean, i mean that

0<= y <= 1-x-z then x+z <= 1 not x+z > 1 as the case that tiny-tim was talkin' about

[STRIKE]Just because [tex]0\leq y \leq 1-x-z[/tex] does not mean that [tex]x+z \leq 1[/tex]. The one does not follow from the other.[/STRIKE] Cancel that.

What tiny-tim is talking about is that if you want to integrate over the entire circle, you have to then subtract the part of the circle where x+z>1.

[EDITED]

 

[LCKurtz]

In problems like this, a picture is worth a thousand words. I plotted y upwards:

The point is that the slanted plane goes "underground" ("ground" means where y = 0) with respect to y, so you can't include the brown area in your domain.

 

[Gabry89]

^then -1 [tex]\leq[/tex] x [tex]\leq[/tex] 0 and the same for z?

I'd use the fact that the plane x+y+z=1 crosses the base of the cylinder which is x²+z²=1, y=0.

What tiny-tim is talking about is that if you want to integrate over the entire circle, you have to then subtract the part of the circle where x+z>1.

i've understood it theoretically but i don't know how to put it into practice: i have to deal separately with the case x+z<=1 and x+z>1 then, is the first case the one that i did in the first post? which gives me - pi/4? so, i just have to do the part related to x+z>1.

 

[SammyS]

thanks, Happy new year too ;)

I thought about that, here's what I've done:

In the xz-plane, you have x² + z² ≤ 1 AND x + z ≤ 1.

x² + z² ≤ 1 is the circle of radius 1, centered at the origin and the interior of this circle. (In cylindrical coordinates, this is ρ ≤ 1 .)

x + z ≤ 1 is the line, x + z = 1, and the half-plane on the origin side of this line. (In cylindrical coordinates, this line has the equation, ρ = 1/(sin(φ)+cos(φ) ) .)

(So, the region described by, x² + z² ≤ 1 AND x + z ≤ 1, is the portion of the above circular region on the origin side of the above line.)

The above line, and circle, intersect at (ρ=1, φ=0) and (ρ=1, φ=π/2).

Therefore, you need to break your integral up.

For 0 ≤ φ ≤ π/2, 0 ≤ ρ ≤ 1/(sin(φ)+cos(φ)).

For π/2 ≤ φ ≤ 2π, 0 ≤ ρ ≤ 1.​

By the way, your integral, [tex]\displaystyle \dots \int_{{{1}\over{\cos\varphi+\sin\varphi}}}^{\ 1} (\dots)\,d\varphi\ \dots[/tex]
integrates only over the region of the circle which should be excluded from the integration.

 

[Gabry89]

^like this then:

 

[SammyS]

^like this then:

Looks good to me. Do the integration & see if the answer checks out.

 

[LCKurtz]

One other note. You can think of the xz domain as a 3/4 circle plus a right triangle. There is no reason to do the right triangle portion in polar coordinates; it only complicates matters. You could integrate the 3/4 circle in polar coordinates and the triangle in rectangular.

 

[Gabry89]

Looks good to me. Do the integration & see if the answer checks out.

Ok, even if the first one looks kinda complicated 'cause i get (talkin' about [tex]\rho)[/tex] [tex]\int[/tex] from 0 to 1/(sin[tex]\varphi[/tex]+cos[tex]\varphi[/tex]) of [tex]\rho[/tex]^2 and besides the same for [tex]\rho[/tex]^3, then is there an easier way to solve these integrals than e.g. calculating the cube?

the second one gives me a part of the answer so it seems right.

 

[SammyS]

Ok, even if the first one looks kinda complicated 'cause i get (talkin' about [tex]\rho)[/tex] [tex]\int[/tex] from 0 to 1/(sin[tex]\varphi[/tex]+cos[tex]\varphi[/tex]) of [tex]\rho[/tex]^2 and besides the same for [tex]\rho[/tex]^3, then is there an easier way to solve these integrals than e.g. calculating the cube?

the second one gives me a part of the answer so it seems right.

If you are allowed to do part of the integration in rectangular coordinates, then using rectangular coordinates for [tex]\displaystyle 0\le\varphi \le\pi/2[/tex] appears to be straight forward.

If not:
It may be helpful to know that [tex] \sin x+\cos x=\sqrt{2}\,\sin\left(x+{{\pi}\over{4}}\right)= \sqrt{2}\,\cos\left(x-{{\pi}\over{4}}\right)\,.[/tex]

So that [tex] {{1}\over{\sin \varphi+\cos \varphi}}=\sqrt{2}\,\csc\left(\varphi+{{\pi}\over{4}}\right)= \sqrt{2}\,\sec\left(\varphi-{{\pi}\over{4}}\right)\,.[/tex]

I've got an appointment to keep. I'll look at this later.

By the way, what part of the answer have you gotten thus far? ... What part of the integral have you done?

 

[Gabry89]

By the way, what part of the answer have you gotten thus far? ... What part of the integral have you done?

i did [tex]\int[/tex]from pi/2 to 2pi [tex]\int[/tex] from 0 to 1 [tex]\int[/tex] 0 to 1-[tex]\rho[/tex]cos[tex]\varphi[/tex]-[tex]\rho[/tex]sin[tex]\varphi[/tex] of [([tex]\rho[/tex]^2)cos[tex]\varphi[/tex]] dy d[tex]\rho[/tex] d[tex]\varphi[/tex]

which gives me -1/3 + 1/8 -(3pi)/16 so a part of the right result ( we can see -(3pi)/16)

 

[SammyS]

It may be helpful to know that [tex] \sin x+\cos x=\sqrt{2}\,\sin\left(x+{{\pi}\over{4}}\right)= \sqrt{2}\,\cos\left(x-{{\pi}\over{4}}\right)\,.[/tex]

So that [tex] {{1}\over{\sin \varphi+\cos \varphi}}=\sqrt{2}\,\csc\left(\varphi+{{\pi}\over{4}}\right)= \sqrt{2}\,\sec\left(\varphi-{{\pi}\over{4}}\right)\,.[/tex]

The above identities don't really help.

i did [tex]\int_{\pi/2}^{2\pi}[/tex] [tex]\int_0^1[/tex] [tex]\int_0^{1-\rho\cos\varphi-\rho\sin\varphi}\rho^2\,\cos\varphi\ dy\ d\rho\ d\varphi[/tex]

which gives me -1/3 + 1/8 -(3pi)/16 so a part of the right result ( we can see -(3pi)/16)

What you got for this part of the integral looks like it's right. BTW: -1/3 + 1/8 = -5/24

^like this then:

Your first integral, [tex]\int_{0}^{\pi/2}\int_0^{1/(\cos\varphi+\sin\varphi)}[/tex] [tex]\int_0^{1-\rho\cos\varphi-\rho\sin\varphi}\rho^2\,\cos\varphi\ dy\ d\rho\ d\varphi[/tex]

will work out pretty nicely. Write the upper limit of the dy integral as [tex]1-\rho(\cos\varphi+\sin\varphi)\,.[/tex]

This will give you:

[tex]\int_{0}^{\pi/2}\int_0^{1/(\cos\varphi+\sin\varphi)}[/tex] [tex]\int_0^{1-\rho(\cos\varphi+\sin\varphi)}\rho^2\,\cos\varphi\ dy\ d\rho\ d\varphi[/tex]

When you plug in for ρ³ and ρ⁴, it will take some algebra & trig, but the answer does work out.

 

[Gabry89]

i'm arrived at this integral: 1/12 *

[tex]
\int_{0}^{\pi/2}\ (cos\varphi)/[cos\varphi+sin\varphi]^3 d\varphi
[/tex]i've seen that it gives me 1/24 using a calculator so I've found the correct result but i just don't know how to get to 1/2 solving the integral.

 

[tiny-tim]

cosφ + sinφ = cos(φ - π/4)√2

(as SammyS said)