Area Under Parametric Curve

Area Under Parametric Curve

In this tutorial we will learn how to find the area under parametric curves. Here the equation of a curve will be given in terms of a parameter. Now we have to find the area of a region bounded by the parametric function under given conditions. If the curve is represented by a cartesian equation y = f(x). Then the area of the region bounded by the non negative function y = f(x), x-axis, and ordinates x = a and x = b is =\int ^{b}_{a}f\left( x\right) dx

Let the equation of a curve y = f(x) is represented parametrically as x = g(t) and y = h(t). Here ‘t’ be a parameter such that α ≤ t ≤ β (Where g(α) = a and h(β) = b).

\begin{aligned}Area\left( A\right) =\int ^{b}_{a}f\left( x\right) dx\\ Let,x=g\left( t\right) \Rightarrow dx=g'\left( t\right) dt\\ \Rightarrow Area\left( A\right) =\int ^{\beta }_{\alpha }h\left( t\right) g'\left( t\right) dt\end{aligned}

The above formula will be used to evaluate the area under parametric curve x = g(t), y = h(t), x-axis and ordinates x = a and x = b. Now we will take a look at some solved example to understand the working of this method.

Example: Compute the area under parametric curve whose equation is given by x = a cost, y = b sint.
Solution: Since the given equation represents a standard ellipse. So here we calculate the area bounded by the upper half of the ellipse and double it in order to find the net area. Using the above formula the area of the region can be given as

\begin{aligned}Area\left( A\right) =2\int ^{0}_{\pi }\left( b\sin t\right) \left( -a\sin t\right) dt\\ =2ab\int ^{\pi }_{0}\sin ^{2}tdt=\pi ab\end{aligned}

Example: Find the area enclosed by the curve x = a sin3t and y = a cos3t.
Solution: Consider the figure 1.1 given in the thumbnail. The required area can be given as

\begin{aligned}Area\left( A\right) =4\int ^{\pi /2}_{0}\left( a^{2/3}-a^{2/3}\sin ^{2}t\right) ^{3/2}3a\left( \sin ^{2}t\right) \left( cost\right) dt\\ =12a^{2}\int ^{\pi /2}_{0}\left( \sin ^{2}t\right) \left( cos^{4}t\right) dt\\ =12a^{2}\left( \dfrac{1\cdot 3\cdot 1\cdot \pi }{6\cdot 4\cdot 2\cdot 2}\right) =\dfrac{\pi \cdot 12a^{2}}{2\cdot 16}=\dfrac{3a^{2}\pi }{8}\end{aligned}

Now attempt at least five questions based on this topic. So this is it from this tutorial. Hoping you people will attempt a few problems based on the topic discussed in this tutorial. In the next tutorial, we will discuss the topic “Area Of The Region Bounded By the Curves Given In their polar Form”.

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