Surface Area¶
Let $f = f(x,y)$ be a continuous function defined on a domain $D$ in the plane. If we let $E$ represent the solid region sitting above (or below) $D$ and bounded between the graph of $f$ and the $xy$-plane, then we know that
$$
\text{Vol}(E) = \int\!\!\!\int_D f(x,y)\, dA.
$$
Furthermore, we have seen that the area of the region $D$ itself can be computed by a double integral,
$$
\text{Area}(D) = \int\!\!\!\int_D 1\, dA.
$$
There is one more important geometric measurement that is naturally given by a double integral: the surface area of the surface $S = \Gamma(f,D)$ given by the graph of $f$ over $D$. In the following video we derive this formula for the surface area, $\text{A}(S)$.
$$
\text{A}(S) = \int\!\!\!\int_D \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}\, dA
$$
Video Lecture
Examples¶
Next, we look at two examples. As always, try them yourself before watching the videos or looking at the answers.
1. Example¶
Compute the surface area of the portion of the paraboloid $z = x^2 + y^2$ that lies below the plane $z = 9$.
Check your answer
$$\frac{\pi}{6}\Big(37\sqrt{37} - 1\Big)$$
Video Solution
2. Example¶
Compute the surface area of the portion of the surface given by the function $f(x,y) = x^2 + 2y$ and sitting above the triangle in the $xy$-plane with vertices $(0,0)$, $(1,0)$, and $(1,1)$.
Check your answer
$$\frac{1}{12}\Big(27 - 5\sqrt{5}\Big)$$
Video Solution
Discussion¶
Questions? You can ask them here.