Question

Find the area of the surface given by z = f(x, y) that lies above the region R. f(x, y) = 3 + 4x3/2 R: rectangle with vertices (0, 0), (0, 5), (2, 5), (2, 0)

Answer 1

It looks like the function is

$f(x,y) = 3 + 4x^{3/2}$

We have

$\dfrac{\partial f}{\partial x} = 6x^{1/2} \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 36x$

$\dfrac{\partial f}{\partial y} = \left(\dfrac{\partial f}{\partial y}\right)^2 = 0$

Then the area of the surface over $R$ is

$\displaystyle \iint_R f(x,y) \, dS = \iint_R \sqrt{1 + 36x + 0} \, dA \\\\ ~~~~~~~~ = \int_0^5 \int_0^2 \sqrt{1+36x} \, dx \, dy \\\\ ~~~~~~~~ = 5 \int_0^2 \sqrt{1+36x} \, dx \\\\ ~~~~~~~~ = \frac5{36} \int_1^{73} \sqrt u \, du \\\\ ~~~~~~~~ = \frac5{36}\cdot \frac23 \left(73^{3/2} - 1^{3/2}\right) = \boxed{\frac5{54} (73^{3/2} - 1)}$