Suppose and are the parametric equations of a curve. Then the area bounded by the curve, the -axis and the ordinates and will be
Now I’ll solve some examples of that.
According to Stroud and Booth (2013)*, “Determine the area of one arch of the cycloid , i.e. find the area of the plane figure bounded by the curve and the -axis between and .”
Now here the given parametric equations of the cycloid are
And I have to find out the area of the plane figure bounded by the curve and the -axis between and . So I’ll start with the formula
First of all, I’ll find out the value of in terms of . As I know the value of is
Next, I’ll differentiate with respect to to get
So that gives
Therefore the area bounded by the curve and the -axis between and is
Then I’ll simplify it to get
As per the standard formulas in trigonometry,
And that means,
So the area will be
Next, I’ll integrate it.
And for that, I’ll use the standard formulas in integration. Thus it will be
Now I’ll substitute the limits to get
As I know, . So I’ll put these values in the equation of the area and simplify it to get
Therefore the area bounded by the curve and the -axis between and is .
Hence I can conclude that this is the answer to the given example.
Now I’ll give another example.
According to Stroud and Booth (2013)*, “The parametric equations of a curve are Find the area under the curve between and .”
Now here the parametric equations of a curve are And I have to find the area under the curve between and . So I’ll start with the formula
First of all, I’ll find out the value of in terms of . As I know the value of is
Next, I’ll differentiate with respect to to get
So that gives
Therefore the area under the curve between and is
Then I’ll simplify it to get
As per the standard formulas in trigonometry,
And that means,
So the area will be
Next, I’ll integrate it.
And for that, I’ll use the standard formulas in integration. Thus it will be
Now I’ll substitute the limits to get
Then I’ll simplify it to get
As I know and . So I’ll put these values in the equation of the area. And that gives
Now I’ll simplify it to get
Thus the area under the curve between and is .