 # Find the area of the part of the plane as shown below that lies in the first octant. 5x + 4y + z =20

This article aims to find the area of the part of the plane that lies in the first octant. The power of double integration is usually used to consider the surface for more general surfaces. Imagine a smooth surface like a blanket blowing in the wind. It consists of many rectangles joined together. More  precisely, let z = f(x,y) be the surface in R3 defined over the region R in the xy plane. cut the xy plane into rectangles.

Each rectangle will protrude vertically onto a piece of surface. The area of the rectangle in the region R is:

$Area=\Delta x \Delta y$

Let $z = f(x,y)$ be a differentiable surface defined over a region $R$. Then its surface is given by

$Area=\iint_{D}(\sqrt(f_{x}^2+f_{y}^2 +1)d_{x}d_{y}$

The plane is given by:

$5x+4y+z=20$

The surface area of an equation of the form $z=f(x, y)$ is calculated by using following formula.

$A=\iint_{D}(\sqrt(f_{x}^2+f_{y}^2 +1)d_{x}d_{y}$

where $D$ is the domain of the integration.

where $f_{x}$ and $f_{y}$ are partial derivatives of $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$.

Let’s determine the integration domain since the plane lies in the first octant.

$x\geq 0, y\geq 0\: and\: z\geq 0$

When we project the $5x+4y+z=20$  on the $xy-plane$, we can see the triangle as $5x+4y=20$.

Hence domain of integration is given by:

$D=(x,y) | (0 \leq x \leq 4), ( 0 \leq y \leq 5-\dfrac{5}{4}x)$

Find partial derivatives $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$.

$\dfrac{\partial z}{\partial x}=-5$

$\dfrac{\partial z}{\partial y}=-4$

Now put these values into the equation of partial fraction to find the area.

$A=\iint_{D}\sqrt((f_{x}^2+f_{y}^2 +1)d_{x}d_{y}$

$A=\int_{0}^{4}\int_{0}^{5-\dfrac{5}{4}x} \sqrt((-5)^2 +(-4)^4+1 )dydx$

$A=\int_{0}^{4}\int_{0}^{5-\dfrac{5}{4}x} \sqrt(42)dydx$

$A=\sqrt(42)\int_{0}^{4} (5-\dfrac{5}{4}x)dx$

$A=\sqrt(42)(5x-\dfrac{5}{4}x^{2})|_{x=0}^{x=4}$

$A=\sqrt(42)(20-10)$

$A=10\sqrt 42\: unit^2$

Therefore, the required area is $10\sqrt 42 \:unit^2$

## Numerical Result

The answer for the area of the part of the plane given as $5x+4y+z=20$ that lies in the first octant is $10\sqrt 42\: unit^2$.

## Example

Determine the area of the part of the plane $3x + 2y + z = 6$ that lies in the first octant.

Solution:

The plane is given by:

$3x+2y+z=6$

The surface area of an equation of the form $z=f(x, y)$ is calculated by using the following formula.

$A=\iint_{D}(\sqrt(f_{x}^2+f_{y}^2 +1)d_{x}d_{y}$

where $D$ is the domain of the integration.

where $f_{x}$ and $f_{y}$ are partial derivatives of $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$.

Let’s determine the integration domain since the plane lies in the first octant.

$x\geq 0, y\geq 0\: and\: z\geq 0$

When we project the $3x+2y+z=6$  on the $xy-plane$, we can see the triangle as $3x+2y=6$.

Hence, the domain of integration is given by:

$D=(x,y) | (0 \leq x \leq 2), ( 0 \leq y \leq 3-\dfrac{3}{2}x)$

Find partial derivatives $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$.

$\dfrac{\partial z}{\partial x}=-3$

$\dfrac{\partial z}{\partial y}=-2$

Now put these values into the equation of partial fraction to find the area.

$A=\iint_{D}\sqrt((f_{x}^2+f_{y}^2 +1)d_{x}d_{y}$

$A=\int_{0}^{2}\int_{0}^{3-\dfrac{3}{2}x} \sqrt((-3)^2 +(-2)^4+1 )dydx$

$A=\int_{0}^{2}\int_{0}^{3-\dfrac{3}{2}x} \sqrt(14)dydx$

$A=\sqrt(14)\int_{0}^{2} (3-\dfrac{3}{2}x)dx$

$A=\sqrt(14)(3x-\dfrac{3}{4}x^{2})|_{x=0}^{x=2}$

$A=\sqrt(14)(6-3)$

$A=3\sqrt 14\: unit^2$

Therefore, the required area is $3\sqrt 14 \:unit^2$

Output for the area of the part of the plane $3x+2y+z=6$ that lies in the first octant is $3\sqrt 14 \:unit^ 2$.