Triple integral in the first octant

Calculus Level 4

What is the value of $\iiint_R \cos (x + y + z)\ dx\ dy\ dz$ , where $R$ is the region bounded by $x \geq 0$ , $y \geq 0$ , $z \geq 0$ and $x + y + z \leq 2\pi$ ?

The answer is 6.2831799999999997652366801048628985881805419921875.

3 solutions

Tunk-Fey Ariawan
Feb 1, 2014

It's easy to notice that region $R$ is bounded by $x=0$ , $\,x=2\pi - y - z$ , $y=0$ , $y=2\pi - z$ , and $\,0\le z \le 2\pi$ . Therefore $\begin{aligned} \int\int\int_R \cos (x+y+z)\,dxdydz &=\int_{z=0}^{2\pi}\int_{y=0}^{2\pi-z}\int_{x=0}^{2\pi-y-z} \cos (x+y+z)\,dxdydz\\ &= \int_{z=0}^{2\pi}\int_{y=0}^{2\pi-z} -\sin (y+z)\,dydz\\ &= \int_{z=0}^{2\pi} (1-\cos z)\,dz\\ &=2\pi\\ &\approx \boxed{6.28} \end{aligned}$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

용철 고
Dec 31, 2013

since x+y+z≤2π and x,y,z is non negative We can set 0≤x≤2π, 0≤y≤2π-x ,0≤z≤x First integral cos(x+y+z) for z in [0,x] Second integral consequence of first for y in [0,2π-x] Finally integral consequence of second for x in [0,2π] Then value of the triple integral is 2π

why is 0 <= z <= x ?

Shyamal Shukla - 7 years, 5 months ago

This solution does not make sense.

I got the answer $2 \sqrt{3} \pi$ by using the substitution $m = x + y + z$ . The area of the plane $m = x + y + z$ inside the first octant is $\frac { \sqrt{3} m^{2} } {2}$ because the region is an equilateral triangle with side length $\sqrt{2} m$ . Therefore, the integral is equal to $\int_{0}^{2 \pi} \frac { \sqrt{3} m^{2} cos(m) } { 2 } dm$ . Integrating by parts gives the answer as $2 \sqrt{3} \pi$ .

Joe Ill - 7 years, 5 months ago

I think this solution is correct, all you need is integral step by step through 3 variables.

About your thinking, i found the mistake. You use the substitution, particularly, your substitution is a layer parallel to the $x + y + z = 2\pi$ plane. The volume of the layer is the product of the base area and the height.

However, $dm$ is not a correct height value because $dm = dx + dy + dz$ , it don't have any dimensions, and can't be considered as the normal of the layer.

Consider we are tracking $dm$ only on x-axis, so $dy, dz = 0, dm = dx$ and the actual height of the layer is $\frac{dm}{\sqrt3}$ , because it's proportional to the height of the pyramid (the height of the pyramid where $x = m$ is $\frac{m}{\sqrt3}$ ).

Therefore, the integral is equal to $\int\limits_0^{2\pi} \frac{m^2 \cos m}{2} dm = 2\pi$

Quang Minh Bùi - 7 years, 5 months ago

Thanks. I thought that the triple integral approach was too hard, so I tried this way. This was pretty easy to overlook.

Joe Ill - 7 years, 5 months ago

So you are assuming that the square root 3 factor cancels the root 3 in the area. I used the same approach but got the area of each plane using heron's formula as (sqrt(3) / 4) * t^2 where t = x + y + z. I am not trying to find any center of mass. All i am doing is integrating a function over a volume and simplifying it to a 1 dimensional integral. If i integrate [cos(t) (sqrt(3/4( t^2)] over 0 to 2*pi, i do not need to find the height and hence i still do not understand why there is no sqrt(3) factor

Sundar R - 7 years, 4 months ago

@Sundar R – Ok, i detected a mistake.. If (x+y+z) = t , then each of the sides of the triangle (last layer of the tetrahedron) will be sqrt(2)*t and not t as i had assumed in my calculations.

Sundar R - 7 years, 4 months ago

Milun Moghe
Jan 13, 2014

$\iiint_{R}cos(x+y+z)dxdydz=\int_{0}^{2\pi}\int_{0}^{2\pi-x}\int_{0}^{2\pi-x-y}cos(x+y+z)dxdydz$ $=\int_{0}^{2\pi}\int_{0}^{2\pi-x}(-sin(x+y))dxdy=\int_{0}^{2\pi}(1-cos(x))dx=2\pi$

integrating should be inside first, outside last.. so must be S(0->2pi) S(0->2pi-z) S(0->2pi-y-z) cos(x+y+z) dx dy dz

Antonio Valente Macarilay - 7 years, 4 months ago

Triple integral in the first octant

The answer is 6.2831799999999997652366801048628985881805419921875.

3 solutions

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