MHB What is the application of Gauss theorem on a pyramid?

[mathmari]

Hey!

Using Gauss theorem I want to calculate $\iint_{\Sigma}f\cdot NdA$, where $\Sigma$ is the closed boundary surface of the pyramid with vertices $(0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0)$ and $f(x,y,z)=(x^2y, 3y^2z, 9xz^2)$ and the perpendicular vectors $N$ to the inside of the pyramid.

From Gauss theorem we have that $$ \iint_{\Sigma}f\cdot N \ dA=\iiint_{\Omega}\nabla\cdot f \ dV$$ We have that $$\nabla \cdot f=\frac{\partial{(x^2y)}}{\partial{x}}+\frac{\partial{(3y^2z)}}{\partial{y}}+\frac{\partial{(9xz^2)}}{\partial{z}}=2xy+6yz+18xz$$

Now we have to find the boundaries for the integral. We have the pyramid

View attachment 7634

So, do we have to find the equation of each line segment of the pyramid? (Wondering)

 

[I like Serena]

Hey!

Using Gauss theorem I want to calculate $\iint_{\Sigma}f\cdot NdA$, where $\Sigma$ is the closed boundary surface of the pyramid with vertices $(0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0)$ and $f(x,y,z)=(x^2y, 3y^2z, 9xz^2)$ and the perpendicular vectors $N$ to the inside of the pyramid.

From Gauss theorem we have that $$ \iint_{\Sigma}f\cdot N \ dA=\iiint_{\Omega}\nabla\cdot f \ dV$$ We have that $$\nabla \cdot f=\frac{\partial{(x^2y)}}{\partial{x}}+\frac{\partial{(3y^2z)}}{\partial{y}}+\frac{\partial{(9xz^2)}}{\partial{z}}=2xy+6yz+18xz$$

Now we have to find the boundaries for the integral. We have the pyramid

So, do we have to find the equation of each line segment of the pyramid? (Wondering)

Hey mathmari! (Smile)

I think we can make it a little easier for ourselves.
From your picture we can see that x and y are both bounded to a square of which the size depends on z.
For $z=0$ we have $0\le x,y \le 1$.
For $z=\frac 12$ we have $0\le x,y \le \frac 12$.
For $z=1$ we have $0\le x,y \le 0$.

Can we deduce what the integral boundaries should be from that? (Wondering)

 

[mathmari]

I think we can make it a little easier for ourselves.
From your picture we can see that x and y are both bounded to a square of which the size depends on z.
For $z=0$ we have $0\le x,y \le 1$.
For $z=\frac 12$ we have $0\le x,y \le \frac 12$.
For $z=1$ we have $0\le x,y \le 0$.

Can we deduce what the integral boundaries should be from that? (Wondering)

So, we have that $0\leq x,y\leq z$ and $0\leq z\leq 1$, right?

Then we get the following:

\begin{align*}\iiint_{\Omega}(2xy+6yz+18xz) \ dV&=\int_0^1\int_0^z\int_0^z(2xy+6yz+18xz) \ dxdydz \\ & = \int_0^1\int_0^z\left [x^2y+6yzx+9x^2z\right ]_{x=0}^z \ dydz \\ & = \int_0^1\int_0^z\left (z^2y+6yz^2+9z^3\right )\ dydz \\ & = \int_0^1\int_0^z\left (7yz^2+9z^3\right )\ dydz \\ & = \int_0^1\left [\frac{7}{2}y^2z^2+9z^3y\right ]_{y=0}^z\ dz \\ & = \int_0^1\left (\frac{7}{2}z^4+9z^4\right )\ dz \\ & = \int_0^1\frac{25}{2}z^4\ dz \\ & = \frac{5}{2}\left [ z^5\right ]_0^1 \\ & = \frac{5}{2}\end{align*}

The given answer is $-1,1$. What have I done wrong? (Wondering)

 

[I like Serena]

Shouldn't it be $0<x,y<1-z$? (Wondering)

 

[mathmari]

Shouldn't it be $0<x,y<1-z$? (Wondering)

Oh yes (Blush) Then we get the folowing:
\begin{align*}\iiint_{\Omega}&(2xy+6yz+18xz) \ dV=\int_0^1\int_0^{1-z}\int_0^{1-z}(2xy+6yz+18xz) \ dxdydz \\ & = \int_0^1\int_0^{1-z}\left [x^2y+6yzx+9x^2z\right ]_{x=0}^{1-z} \ dydz \\ & = \int_0^1\int_0^{1-z}\left [(1-z)^2y+6yz(1-z)+9(1-z)^2z\right ] \ dydz \\ & = \int_0^1\int_0^{1-z}\left [(1-2z+z^2)y+6yz(1-z)+9(1-2z+z^2)z\right ] \ dydz \\ & = \int_0^1\int_0^{1-z}\left [y-2zy+z^2y+6yz-6yz^2+9z-18z^2+9z^3\right ] \ dydz \\ & = \int_0^1\int_0^{1-z}\left [y+4yz-5yz^2+9z-18z^2+9z^3\right ] \ dydz \\ & = \int_0^1\left [\frac{y^2}{2}+2y^2z-\frac{5}{2}y^2z^2+9zy-18z^2y+9z^3y\right ]_{y=0}^{1-z} \ dz \\ & = \int_0^1\left [\frac{(1-z)^2}{2}+2(1-z)^2z-\frac{5}{2}(1-z)^2z^2+9z(1-z)-18z^2(1-z)+9z^3(1-z)\right ] \ dz \\ & = \int_0^1\left [\frac{1-2z+z^2}{2}+2(1-2z+z^2)z-\frac{5}{2}(1-2z+z^2)z^2+9z-9z^2-18z^2+18z^3+9z^3-9z^4\right ] \ dz \\ & =\int_0^1\left [\frac{1}{2}-z+\frac{z^2}{2}+2z-4z^2+2z^3-\frac{5}{2}z^2+5z^3-\frac{5}{2}z^4+9z-27z^2+27z^3-9z^4\right ] \ dz \\ & = \int_0^1\left [\frac{1}{2}-\frac{23}{2}z^4+10z-33z^2+34z^3\right ] \ dz \\ & = \left [\frac{1}{2}z-\frac{23}{10}z^5+5z^2-11z^3+\frac{34}{4}z^4\right ]_0^1 \\ & = \frac{1}{2}-\frac{23}{10}+5-11+\frac{34}{4}\\ & = \frac{7}{10}\end{align*}

Is everything correct? (Wondering)

Is at the given answer a typo or have I done something wrong? (Wondering)

 

[I like Serena]

Is everything correct?

Is at the given answer a typo or have I done something wrong?

Everything looks correct to me. (Nod)
And W|A confirms that your calculation of the integral is correct as well.
So I think there is indeed a mistake in the given answer. (Thinking)

 

[mathmari]

Everything looks correct to me. (Nod)
And W|A confirms that your calculation of the integral is correct as well.
So I think there is indeed a mistake in the given answer. (Thinking)

Great! Thank you! (Happy)

 

[mathmari]

What would we do if the perpendicular vectors $N$ would direct to the outside of the pyramid? (Wondering)

 

[I like Serena]

What would we do if the perpendicular vectors $N$ would direct to the outside of the pyramid?

Oh! I overlooked that before! (Wait)

Gauss's theorem assumes that the perpendicular vectors are oriented outward.
From wiki:

The closed manifold ∂V is quite generally the boundary of V oriented by outward-pointing normals, and n is the outward pointing unit normal field of the boundary ∂V.​

Since the problem statement says that they are oriented inward, it means that we have to take the opposite. (Thinking)

 

[mathmari]

Oh! I overlooked that before! (Wait)
Gauss's theorem assumes that the perpendicular vectors are oriented outward.
Since the problem statement says that they are oriented inward, it means that we have to take the opposite.

So, we have to write a "-" everywhere and so we get $-0.7$, right? (Wondering)

 

[I like Serena]

So, we have to write a "-" everywhere and so we get $-0.7$, right?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Thank you! (Sun)