Find surface area that lies above a triangle

Question

Determine the area of the part of the surface $z=2 + 7x + 3y^2$ that lies above the triangle with vertices $(0,0)$, $(0,8)$, and $(14,8)$.

I do not know what formula to use to attempt this problem!

Answer 1

Note that the region in the xy-plane is bounded by the lines x = 0, y = 1, and y = x/2. Write this as x = 0 to x = 2y for y in [0, 1].

Using Cartesian Coordinates, A = ∫∫ √[1 + (∂z/∂x)^2 + (∂z/∂y)^2] dA ...= ∫(y = 0 to 1) ∫(x = 0 to 2y) √[1 + 3^2 + (4y)^2] dx dy ...= ∫(y = 0 to 1) 2y √(10 + 16y^2) dy ...= ∫(w = 10 to 26) 2y √w * (dw/(32y)), letting w = 10 + 16y^2 ...= (1/16) ∫(w = 10 to 26) w^(1/2) dw ...= (1/16) * (2/3)w^(3/2) {for w = 10 to 26} ...= (1/24) (26^(3/2) - 10^(3/2)).

I hope this helps!

Answer 2

For that (right) triangle, we are bounded by $x \in [0,14]$, $y \in [0,(4/7) x]$. The surface area above this triangle is given by

$$\begin{align}\int_0^8 dx \: \int_0^{(4/7) x} dy \: \sqrt{1+\left ( \frac{\partial z}{\partial x}\right )^2+\left ( \frac{\partial z}{\partial y}\right )^2} &= \int_0^8 dx \: \int_0^{(4/7) x} dy \: \sqrt{49 + 36 y^2}\end{align}$$

The inner integral may be performed with either a trig or a sinh substitution; you should be able to prove to yourself that the inner integral is equal to

$$\frac{49}{12} \left [\text{arcsinh}{\left( \frac{24 x}{49}\right )} + \frac{24 x}{49} \sqrt{1+\left(\frac{24 x}{49}\right)^2}\right] $$

The surface area is then the integral of the above expression from $0$ to $8$. You may show by integrating by parts that

$$\int_0^b du\:\text{arcsinh}(a u) = b\, \text{arcsinh}(a b) - \frac{1}{a} \left ( \sqrt{1+a^2\, b^2}-1\right) $$

which is useful in deriving your final result.