Question

Find the area of the surface. The part of the plane 2x + 13y + z = 26 that lies in the first octant

Answer 1

Answer:

Surface Integral

Step-by-step explanation:

If the surface is defined as $\overrightarrow{r} (x,y)$, then integrating the area of one patch $|\overrightarrow{r}_{x} X \overrightarrow{r}_{y}|$ over the whole domain D will give the surface area.

In this case we can write:

$\overrightarrow{r}(x,y) = (x,y,26-2x-13y)$

$\overrightarrow{r}(x,y) \Rightarrow  |\overrightarrow{r}_{x} X \overrightarrow{r}_{y}|$

$|\overrightarrow{r}_{x} X \overrightarrow{r}_{y}| = \sqrt{fx^2}+fy^2}+1 } =\sqrt{(-2)^2+(-13)^2+1}}$

$\int\int\limits_S  dS = \int\int\limits_D \sqrt{184} dA$

To see where the plane intersects the first octant, look for the intercepts with the x, y and z axes:

(2,0,0),(0,13,0)(0,0,26)

So the domain D is the triangle in the first quadrant of the xy plane bounded by (0,0) (2,0) and (0,13). We could compute this using geometry:

$\int\int_{D} \sqrt{184} dA= \sqrt{184} *\frac{1}{2} (2)(13) = 26*2*\sqrt{46}=52\sqrt{46}$

Answer 2

The area of the surface that lies in the first octant is approximately 171.471 square units.

How to find the area of the surface of a plane that lies in the first octant

Let be $2\cdot x + 13\cdot y + z = 26$, the area of the surface that lies in the first octant represents a triangle. The vertices of such triangle are located at $(x_{1},y_{1},z_{1}) = (0, 0, 26)$, $(x_{2}, y_{2}, z_{2}) = (0, 2, 0)$ and $(x_{3}, y_{3}, z_{3}) = (13, 0, 0)$. Now we must find the length of the sides of the triangle:

$a = \sqrt{y_{2}^{2}+z_{1}^{2}}$   (1) (yz-Plane)

$b = \sqrt{x_{3}^{2}+z_{1}^{2}}$   (2)  (xz-Plane)

$c = \sqrt{x_{3}^{2}+y_{2}^{2}}$   (3)  (xy-Plane)

Then, the side lengths are the following: $a = 2\sqrt{170}$, $b = 13\sqrt{5}$, $c = \sqrt{173}$.

The area of the surface ($A$) is found by the following two formulas:

$A = \sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}$   (4)

$s = \frac{a+b+c}{2}$

Where $s$ is the semiperimeter of the triangle.

If we know that $a = 2\sqrt{170}$, $b = 13\sqrt{5}$ and $c = \sqrt{173}$, then the area of the surface is:

$s = \frac{2\sqrt{170}+13\sqrt{5}+\sqrt{173}}{2}$

$s\approx 34.149$

$A = \sqrt{34.149\cdot (34.149-2\sqrt{170})\cdot (34.149-13\sqrt{5})\cdot (34.149-\sqrt{173})}$

$A \approx 171.471$

The area of the surface that lies in the first octant is approximately 171.471 square units. $\blacksquare$

To learn more on surfaces, we kindly invite to check this verified question: https://brainly.com/question/2835293