Question

Find the area of the part of the plane 5x 3y z = 15 that lies in the first octant.

Answer 1

This part of the plane lies above a triangle with boundaries $x=0$ and $y=0$ along the coordinate axes, as well as the line

$z=0 \implies 5x + 3y = 15 \implies y = \dfrac{15 - 5x}3$

When $y=0$, we have $15-5x=0\implies x=3$. So this triangle is the set

$T = \left\{(x,y)\in\Bbb R^2 ~:~ 0\le x\le3 \text{ and } 0\le y\le\dfrac{15-5x}3\right\}$

Also, when $x=0$, we have $y=\frac{15}3=5$. So the triangle has length 3 and width 5, hence area 1/2•3•5 = 15/2.

Let $z=f(x,y) = 15 - 5x - 3y$. Then the area of the plane over $T$ is

$\displaystyle \iint_T dA = \iint_T \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy$

We have

$\dfrac{\partial f}{\partial x} = -5 \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 25$

$\dfrac{\partial f}{\partial y} = -3 \implies \left(\dfrac{\partial f}{\partial y}\right)^2 = 9$

$\implies\displaystyle \iint_T dA = \sqrt{35} \iint_T dx\,dy = \boxed{\frac{15\sqrt{35}}2}$

since the integral

$\displaystyle \iint_TdA$

is exactly the area of $T$, 15/2.