\(\int\limits_{C}y^3\, ds\), \(C: x=t^3,\, y=t,\, 0\leq t\leq 5\).
This question aims to find the line integral given the parametric equations of the curve.
A curve represents the path of a point that is moving continuously. An equation is typically used to generate such a path. The term can also refer to a straight line or a series of linked line segments. A path that repeats itself is called a closed curve, enclosing one or more regions. Ellipses, polygons, and circles, are some examples of this, and open curves with infinite length include hyperbolas, parabolas, and spirals.
An integral of a function along a curve or path is said to be a line integral. Let $s$ be the sum of all the arc lengths of a line. A line integral takes two dimensions and combines them into $s$ and then integrates the functions $x$ and $y$ over the line $s$.
If a function is defined on a curve, the curve can be split up into small line segments. All the products of function value on the segment by the length of line segments can be added and a limit is taken as the line segments tends to zero. This refers to a quantity known as a line integral, which can be defined in two, three, or higher dimensions.
Expert Answer
The line integral over a curve can be defined as:
$\int\limits_{C}f(x,y)\,ds=\int\limits_{a}^{b}f(x(t),y(t))\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt$ (1)
Here, $f(x,y)=y^3$ and $\vec{r}(t)=\langle x(t), y(t) \rangle=\langle t^3, t \rangle$
Also, $\vec{r}'(t)=\langle 3t^2, 1 \rangle$
Now, $ds=|\vec{r}'(t)|\,dt=\sqrt{\left(3t^2\right)^2+\left(1\right)^2}\,dt$
$ds=\sqrt{9t^4+1}\,dt$
Therefore, form (1):
$\int\limits_{C}f(x,y)\,ds=\int\limits_{0}^{3}t^3\cdot \sqrt{9t^4+1}\,dt$
Using integration by substitution:
Let $u=9t^4+1$ then $du=36t^3\,dt$ or $t^3\,dt=\dfrac{du}{36}$
For limits of integration:
When $t=0\implies u=1$ and when $t=3\implies u=730$
So, $\int\limits_{0}^{3}t^3\cdot \sqrt{9t^4+1}\,dt=\int\limits_{1}^{730}\sqrt{u}\,\dfrac{du}{36}$
$=\dfrac{1}{36}\int\limits_{1}^{730}\sqrt{u}\,du$
$=\dfrac{1}{36}\int\limits_{1}^{730}u^{\frac{1}{2}}\,du$
$=\dfrac{1}{36}\left[\dfrac{u^{\frac{3}{2}}}{\dfrac{3}{2}}\right]_{1}^{730}$
$=\dfrac{1}{54}\left[u^{\frac{3}{2}}\right]_{1}^{730}$
Apply limits of integration:
$=\dfrac{1}{54}\left[(730)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]$
$=\dfrac{1}{54}[19723.51-1]$
$=\dfrac{1}{54}[19722.51]$
$=365.23$
Graph of the given curve along with its surface area
Example 1
Evaluate the line integral $\int\limits_{C}2x^2\,ds$, where $C$ is the line segment from $(-3,-2)$ to $(2,4)$.
Solution
Since the line segment from $(-3,-2)$ to $(2,4)$ is given by:
$\vec{r}(t)=(1-t)\langle -3,-2\rangle+t\langle 2,4\rangle$
$\vec{r}(t)=\langle -3+5t,-2+6t\rangle$, where $0\leq t\leq 1$ for the line segments from $(-3,-2)$ to $(2,4)$.
From above, we have the parametric equations:
$x=-3+5t$ and $y=-2+6t$
Also, $\dfrac{dx}{dt}=5$ and $\dfrac{dy}{dt}=6$
Therefore, $ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt$
$=\sqrt{(5)^2+(6)^2}=\sqrt{61}$
And so, $\int\limits_{C}2x^2\,ds=\int\limits_{0}^{1}2(-3+5t)^2(\sqrt{61})\,dt$
$=2\sqrt{61}\int\limits_{0}^{1}(-3+5t)^2\,dt$
$=\dfrac{2\sqrt{61}}{5}\left[\dfrac{(-3+5t)^3}{3}\right]_{0}^{1}$
Apply limits of integration as:
$=\dfrac{2\sqrt{61}}{15}\left[(-3+5(1))^3-(-3+5(0))^3\right]$
$=\dfrac{2\sqrt{61}}{15}\left[8-(-27)\right]$
$=\dfrac{2\sqrt{61}}{15}\left[35\right]$
$=36.44$
Example 2
Given $C$ as the right half of the circle $x^2+y^2=4$ in an anti-clockwise direction. Calculate $\int\limits_{C}xy\,ds$.
Solution
Here, the parametric equations of the circle are:
$x=2\cos t$ and $y=2\sin t$
Since $C$ is the right half of the circle in anti-clockwise direction, therefore, $-\dfrac{\pi}{2}\leq t\leq \dfrac{\pi}{2}$.
Also, $\dfrac{dx}{dt}=-2\sin t$ and $\dfrac{dy}{dt}=2\cos t$
And so, $ds=\sqrt{(-2\sin t)^2+(2\cos t)^2}\,dt$
$ds=\sqrt{4\sin^2t+4\cos^2t}\,dt=2\,dt$
$\int\limits_{C}xy\,ds=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(2\cos t)(2\sin t)(2)\,dt$
$=8\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin t (\cos t\,dt)$
$=8\left[\dfrac{\sin^2t}{2}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$
$=4\left[\left(\sin \left(\dfrac{\pi}{2}\right)\right)^2-\left(\sin \left(-\dfrac{\pi}{2}\right)\right)^2\right]$
$=4[1-1]$
$=0$
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