** 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**

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

**Expert Answer**

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 d**omain 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 d**omain 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 $3\sqrt 14 \:unit^ 2$.**