Find surface area that lies above a triangle 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!
 A: 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.
A: 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!
