Arc Length and Surface Area in Parametric Form | Free Calculus 2 Lesson (4 College Credits) | UPI Study

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Arc Length and Surface Area in Parametric Form

Calculus 2 · Chapter 9: Parametric Equations & Polar CoordinatesSection 9.3 · lesson 47 of 549 min · free

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In this lesson

The same ds, now driven by a parameter.

  • Rewrite ds with both legs in terms of t
  • Measure curves no function of x can describe
  • Attach 2\pi r to get a surface of revolution
  • Guard against retracing the curve

Transcript

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MAT-265 - Calculus 2 · Section 9.3

Arc Length and Surface Area

Section four point four built the arc length element from a right triangle. Legs d x and d y, hypotenuse d s. And that reasoning never depended on how the curve was described. So for a parametric curve, you simply express each leg through the parameter, and factor d t out of the radical. The same idea, in new clothing. And it unlocks curves that no function of x could ever handle.

The Element

Both Legs Through the Parameter

Start from the element. d s squared equals d x squared plus d y squared. Now write each leg in terms of t. d x equals x prime of t, d t. And d y equals y prime of t, d t. Substitute and factor the common d t out of the square root. You get d s equals root of d x d t squared, plus d y d t squared — all times d t. Integrating that gives the arc length theorem. Let a curve be given by x of t and y of t, for t between alpha and beta. Where the derivatives are continuous, and the curve is traversed exactly once as t increases. Then the length L is the integral from alpha to beta, of that radical, d t. Now — that requirement about being traversed exactly once is essential. Not a technicality. If the parameter interval causes the curve to be retraced, the integral counts the retraced portion again. And overstates the true length.

Example 1

Test It Where the Answer Is Known

Before applying the formula to an unfamiliar curve, it's worth testing it where the answer is already known. A circle of radius three, centered at the origin, is x equals three cosine t, y equals three sine t, for t from zero to two pi. Differentiate each coordinate. d x d t is minus three sine t. d y d t is three cosine t. Square each and add. Squaring removes the negative sign from the first. You get nine sine squared t, plus nine cosine squared t. Which is nine, times sine squared plus cosine squared. The Pythagorean identity collapses the bracket to one. Leaving the constant nine. So the square root is three. And the integral becomes three t, evaluated from zero to two pi. That's six pi. About eighteen point eight five. And the circumference of a circle of radius three is two pi times three. Six pi. An exact match. The formula passes its first test. Now note the role of the parameter interval. Had t run from zero to four pi instead, the integral would have returned twelve pi. Double the true circumference. Because the circle would have been traversed twice.

Example 2

A Curve No Function Can Describe

The real power of the formula appears on curves that cannot be written as y equals f of x. Such curves are common in mechanics — where the position of a point is naturally a function of time, rather than of horizontal displacement. An astroid is the curve traced by a point on a small circle rolling inside a larger one. Its first-quadrant portion is x equals cosine cubed t, y equals sine cubed t, for t from zero to pi over two. Differentiate using the chain rule. For the first coordinate, the outer function is a cube and the inner is cosine t, whose derivative is minus sine t. So d x d t is three cosine squared t, times minus sine t. Which is minus three cosine squared t sine t. Similarly d y d t is three sine squared t cosine t. Square each and add. You get nine cosine to the fourth t sine squared t, plus nine sine to the fourth t cosine squared t. Both terms share the common factor nine sine squared t cosine squared t. Factor it out. The first term leaves cosine squared t, the second leaves sine squared t. And that bracket collapses to one, by the Pythagorean identity again. So the whole thing is nine sine squared t cosine squared t. Take the square root. And note that on zero to pi over two, both sine and cosine are nonnegative — so no absolute value bars are needed. You get three sine t cosine t. Now apply the double-angle identity. Sine two t equals two sine t cosine t. So sine t cosine t is one half sine two t. The integrand becomes three halves, sine two t. Integrate. The antiderivative of sine two t is minus one half cosine two t. Evaluate at both limits. At t equals pi over two, cosine two t is cosine pi, which is minus one. At t equals zero, it's one. So L equals minus three quarters, times minus one minus one. Which is three halves. One point five. The arc has length exactly one point five units. And no representation of this curve as y equals f of x was needed at any stage.

Figure 9.3

The Two Curves

Here are the two curves. On the left, the astroid arc. Starting at one comma zero when t is zero, and ending at zero comma one when t is pi over two. Notice its shape — it curves inward, concave toward the origin. That's the signature of an astroid. And it fails the vertical line test nowhere here, but the full astroid — all four quadrants — certainly would. On the right, the circle from Example one. Whose known circumference gave us confidence in the formula before we applied it to anything unfamiliar.

Surface Area

Attach a Circumference to the Element

Revolving a curve about an axis sweeps out a surface. Section four point five developed the surface area formula by treating each infinitesimal band as a frustum of a cone. With slant height d s, and circumference two pi times the radius. And that reasoning is likewise independent of how the curve is described. Only the expressions for the radius, and for d s, need translating into parametric language. When the curve is revolved about the x-axis, the radius of the band swept out by a point is its vertical distance from that axis. Namely y of t. When the revolution is about the y-axis, the radius is the horizontal distance, x of t, instead. So — revolving about the x-axis, where y of t is nonnegative throughout, generates a surface of area: the integral of two pi y of t, times the same radical, d t. And revolving about the y-axis, where x of t is nonnegative, uses x of t as the radius instead. In each case the radical is identical to the one in the arc length theorem. The only new ingredient is the factor two pi times the radius. Let's do one. Find the area of the surface generated by revolving the astroid arc about the x-axis. Since the arc lies in the first quadrant, sine cubed t is nonnegative on the whole interval. So the x-axis formula applies. And the radical was already computed in Example two. It's three sine t cosine t. Substitute that, and the radius sine cubed t, into the surface area integral. You get the integral of two pi sine cubed t, times three sine t cosine t. Which is six pi, times sine to the fourth t, cosine t. That integrates by substitution to sine to the fifth t, over five. Evaluate from zero to pi over two. Sine of pi over two is one, so you get one fifth. The surface area is six pi over five.

Key Takeaways

What to Carry Forward

Five takeaways. The arc length element is unchanged from Chapter Four. Only its legs are now expressed through t — which lets d t factor cleanly out of the radical. The length formula measures curves that no function y equals f of x can describe. Which is the whole point. The curve must be traversed exactly once. Otherwise the integral counts retraced portions again, and overstates the length. Surface area attaches two pi r to the same radical. Radius y of t for the x-axis, x of t for the y-axis — with the radius nonnegative throughout. And watch for Pythagorean and double-angle identities. Workable parametric problems are usually engineered so the radical collapses. Next up is section nine point four — the Polar Coordinate System. A completely different way to locate a point in the plane.

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FAQ

Questions students ask us

About Arc Length and Surface Area in Parametric Form

What is the arc length formula for a parametric curve?
The arc length of a parametric curve x(t), y(t) for \alpha \le t \le \beta is L = \int_\alpha^\beta \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt. It comes from the right-triangle element ds^2 = dx^2 + dy^2. Substitute dx = x'(t)\,dt and dy = y'(t)\,dt, then factor dt out of the radical. The derivatives must be continuous, and the curve must be traversed exactly once as t increases.
How do you find the surface area of revolution for a parametric curve?
To find the surface area, multiply the parametric arc length element by the circumference 2\pi r and integrate. About the x-axis, the radius is y(t), so S = \int_\alpha^\beta 2\pi y(t)\sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt, which requires y(t) \ge 0. About the y-axis, use x(t) as the radius instead, which requires x(t) \ge 0. The radical is identical to the one in the arc length formula.
How do you find the arc length of an astroid using parametric equations?
The first-quadrant astroid arc x = \cos^3 t, y = \sin^3 t, 0 \le t \le \pi/2, has length 3/2. Its derivatives are -3\cos^2 t \sin t and 3\sin^2 t \cos t. Squaring, adding, and using \sin^2 t + \cos^2 t = 1 gives 9\sin^2 t\cos^2 t. The square root is 3\sin t\cos t = \tfrac{3}{2}\sin 2t, which integrates to 3/2. Revolving this arc about the x-axis gives a surface area of 6\pi/5.
Why does my parametric arc length answer come out too big?
Your parametric arc length is probably too big because the parameter interval makes the curve retrace itself, so the integral counts the retraced portions again. For example, the circle x = 3\cos t, y = 3\sin t gives the correct circumference 6\pi on 0 \le t \le 2\pi. On 0 \le t \le 4\pi it gives 12\pi because the circle is traversed twice. Always check that the curve is traced exactly once.

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