Surface Area With Double Integrals (How-To w/ Step-by-Step Examples!)

Surface area is crucial to knowing…

Jenn (B.S., M.Ed.) of Calcworkshop® teaching surface area with double integrals

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

How much paint we need to cover a wall.
How much carpet we need to cover a floor.
Or how much fabric we need to make an article of clothing.

While we are familiar with how to calculate surface area using basic geometric formulas…

…what about finding area of surfaces associated with plane curves?

This is where the double integral and its trusty sidekick arc length come in handy.

Let’s find out how.

Surface Area w/ Double Integrals

Remember how we learned about arc length over an interval in single variable calculus and then extended that idea to find the surface area of a solid of revolution?

Well, now we will take both concepts and adapt them to finding surface area over a region for a function of two variables.

Here’s the big idea without getting lost in the weeds

Suppose $S$ is a surface with equation $z = f (x, y)$ above the xy-plane with region $R$ , where $f$ has continuous partial derivatives.

Then if we divide $R$ into $n$ rectangular subregions, which have corresponding regions on the surface, we create an infinite number of tangent planes or tangent parallelograms to the surface.

And if we determine the area of these $n$ number of tangent planes, the total surface area is the sum of all of these tiny tangent plane areas.

Thus, the area of the surface $S$ with equation $z = f (x, y)$ with continuous partial derivatives is:

$A (S) = \iint_{R} \sqrt{1 + {(f_{x})}^{2} + {(f_{y})}^{2}} d A$

And I hope this formula looks familiar to you, as it’s strikingly similar to the arc length formula from single-variable calculus.

Example

For example, let’s determine the surface area of the plane S given by $f (x, y) = 9 + x + 2 y$ over the rectangle $R = {(x, y) : 0 \leq x \leq 3, 0 \leq y \leq 2}$ .

Projection Surface Area Plane

First, we will calculate our first partial derivatives.

$\frac{\partial f}{\partial x} = f_{x} = 1 and \frac{\partial f}{\partial y} = f_{y} = 2$

Next, we will substitute our partials into our surface area for double integrals formula and solve.

$A (S) = \iint_{R} \sqrt{1 + {(f_{x})}^{2} + {(f_{y})}^{2}} d A$

$A (S) = \int_{0}^{3} \int_{0}^{2} \sqrt{1 + (1)^{2} + (2)^{2}} d y d x$

$= \int_{0}^{3} \int_{0}^{2} (\sqrt{6}) d y d x$

$= \int_{0}^{3} ([\sqrt{6} y]_{0}^{2}) d x$

$= \int_{0}^{3} 2 \sqrt{6} d x = {2 \sqrt{6} x |}_{0}^{3} = 6 \sqrt{6}$

Simple.

But what about finding the surface area for regions that are not rectangular but general?

Never fear. This formula can handle non-rectangular regions too.

Example

Let’s look at another example.

Determine the surface area of the hyperbolic paraboloid $f (x, y) = 9 + x^{2} + y^{2}$ over the circular disc $R = {(x, y) : 0 \leq x^{2} + y^{2} \leq 4}$ .

Projection Hyperbolic Paraboloid

Like before, the first thing we will do is find the first order partials of our function $f$ .

$\frac{\partial f}{\partial x} = f_{x} = 2 x and \frac{\partial f}{\partial y} = f_{y} = 2 y$

Next, we will substitute our partial derivatives into our surface area formula.

$A (S) = \iint_{R} \sqrt{1 + (2 x)^{2} + (2 y)^{2}} d A = \iint_{R} \sqrt{1 + 4 x^{2} + 4 y^{2}} d A$

Now, this integrand isn’t as “nice” as our previous example and will be quite a bear to calculate in its current form.

So, what do we do?

Well, we notice that our region $R$ is a disk, so maybe it would be easier if we changed everything to polar coordinates?

And that’s precisely what we are going to do!

Our region $R$ is a circle, which means we can rewrite our domain as follows:

$\begin{matrix} x^{2} + y^{2} = 4 \\ r^{2} = 4 \\ r = 2 \end{matrix}$

And since the boundary of our circle is a full rotation, we can conclude $0 \leq θ \leq 2 π$ .

Now all that is left is to rewrite our integrand in polar coordinates…

…but don’t forget to add the Jacobian determinant $r$ !

$A (S) = \iint_{R} \sqrt{1 + 4 (x^{2} + y^{2})} d A = \iint_{R} (\sqrt{1 + 4 (r^{2})}) r d r d θ$

Perfect. Now it’s time to substitute our limit values and evaluate the iterated integral! Note that we will need to use u-substitution for our inner integral.

$A (S) = \int_{0}^{2 π} \int_{0}^{2} (\sqrt{1 + 4 r^{2}}) r d r d θ$

$= \int_{0}^{2 π} ({[\frac{1}{12} {(1 + 4 r^{2})}^{3 / 2}]}_{0}^{2}) d θ$

$= \int_{0}^{2 π} (\frac{(17)^{3 / 2} - 1}{12}) d θ$

$= \frac{((17)^{3 / 2} - 1) π}{6} \approx 36.177$

Not too bad, right?

But I must warn you of a few things.

First, the integration involved in computing these integrals by hand can be tedious and take a little bit of ingenuity (i.e., changing to polar coordinates or using u-substitution or inverse hyperbolic trig functions). So, take heart and be savvy.
Secondly, there may be times when we can’t evaluate the antiderivative without using a calculator or computer software. So, have your technology at the ready.
And finally, we will be dealing with finding the surface area of more general surfaces, called parametric surfaces, where our surface is defined by a vector-valued function in a future lesson. So, get excited about having more surface area in your future.

But let’s not wait any longer. Let’s jump right into our lesson and discover how to combine arc length and tangent planes to find the surface area using double integrals!

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

Surface-Area-using-Double-Integrals-Example

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