Examples of Line Integrals - Parametrizing Curves (Example 4)

Sometimes you may be asked to do a line integral on 3-dimensional curves, you just have to parametrize the curves and create the integral.

Find the line integral \(\int_C^{} {{y^3}ds} \) given that the curve is described by C: \(x = {t^3}\), \(y = t\), for \(0 \le t \le 2\).

Use the following formula, which relates ds to dx, dy, dz:

\[ds = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt\]

And we plug in the derivatives for the equations we were given.

\[ds = \sqrt {{{\left( {3{t^2}} \right)}^2} + {{(1)}^2}} dt\]

\[ds = \sqrt {9{t^4} + 1} dt\]

So now we can make the integral. We use the boundaries given in the question, t goes from 0 to 2. We substitute in y=t, and we also use the relation we made between ds and dx, dy, dz. Let's call the integral "I".

\[I = \int_0^2 {{t^3}\sqrt {9{t^4} + 1} } dt\]

And we can use regular substitution to solve this integral. One of the easy ways of knowing you need to use substitution is if you see variables that have powers one apart.

\[u = 9{t^4} + 1\]

\[du = 36{t^3}dt\]

\[\frac{1}{{36}}du = {t^3}dt\]

\[u(0) = 1\]

\[u(2) = 9{(2)^4} + 1 = 9(16) + 1 = 145\]

\[I = \frac{1}{{36}}\int_1^{145} {\sqrt u } du\]

\[I = \frac{1}{{36}} \cdot \frac{2}{3} \cdot \left[ {{{(145)}^{\frac{3}{2}}} - {{(1)}^{\frac{3}{2}}}} \right]\]

And we can clean it up a bit. One to the power of any real number is just one. And we can simplify the fractions to get a final answer.

\[I = \frac{{145\sqrt {145} - 1}}{{54}}\]

Please see our other worked-out line integral examples:

Line Integrals (Ex1)

Line Integrals (Ex2)

Line Integrals Multiple Line Segments (Ex3)

Alternatively return to the line integral example hub page:

Line Integrals

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