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Area with parametric equation

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I start read about area of parametric equation and got some problem understanding.
I got two question. here is a link

1. Does it mather if we say \(\displaystyle x=f(t)\) and \(\displaystyle y=g(t)\) on \(\displaystyle a \leq x \leq b\)
Does this both formula works?
\(\displaystyle \int_\alpha^\beta g(t)f'(t)dt\) and \(\displaystyle \int_\alpha^\beta g'(t)f(t)dt\)

2. Could someone give me an exemple on that "Area Under Parametric Curve, Formula II"


\(\displaystyle \int_\beta^\alpha f(t)g(t)dt\)

Regards,
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hello MHB,
I start read about area of parametric equation and got some problem understanding.
I got two question. here is a link

1. Does it mather if we say \(\displaystyle x=f(t)\) and \(\displaystyle y=g(t)\) on \(\displaystyle a \leq x \leq b\)
Does this both formula works?
\(\displaystyle \int_\alpha^\beta g(t)f'(t)dt\) and \(\displaystyle \int_\alpha^\beta g'(t)f(t)dt\)
You can do that, but you'll be tracing a different area.
In the first case you're tracing the curve by changing its x-coordinate and you'll get the area between the curve and the x-axis.
In the second case you're changing its y-coordinate and you'll get the area between the curve and the y-axis.

2. Could someone give me an exemple on that "Area Under Parametric Curve, Formula II"

\(\displaystyle \int_\beta^\alpha f(t)g(t)dt\)
Sure.
Pick the same example as in the article, but with $x=-6(\theta-\sin \theta)$.

The point is that the curve is supposed to be traced from the lowest x coordinate to the highest x coordinate.
If the definition of f(t) means it starts at the highest x coordinate and ends at the lowest x coordinate, you should swap the integral boundaries to compensate.
 

Petrus

Well-known member
Feb 21, 2013
739
You can do that, but you'll be tracing a different area.
In the first case you're tracing the curve by changing its x-coordinate and you'll get the area between the curve and the x-axis.
In the second case you're changing its y-coordinate and you'll get the area between the curve and the y-axis.



Sure.
Pick the same example as in the article, but with $x=-6(\theta-\sin \theta)$.

The point is that the curve is supposed to be traced from the lowest x coordinate to the highest x coordinate.
If the definition of f(t) means it starts at the highest x coordinate and ends at the lowest x coordinate, you should swap the integral boundaries to compensate.
Hello I like Serena,
I am glad that I got a responed, thank you!
I got one question about
You can do that, but you'll be tracing a different area.
In the first case you're tracing the curve by changing its x-coordinate and you'll get the area between the curve and the x-axis.
In the second case you're changing its y-coordinate and you'll get the area between the curve and the y-axis.
In the problem it never mention what area they want, I read also from my book and it use the one with \(\displaystyle \int_\alpha^\beta g(t)f'(t)dt\) but on example it never mention what area tracing or I am confused?

Regards,
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
In the problem it never mention what area they want, I read also from my book and it use the one with \(\displaystyle \int_\alpha^\beta g(t)f'(t)dt\) but on example it never mention what area tracing or I am confused?
Paul writes:
We will do this in much the same way that we found the first derivative in the previous section. We will first recall how to find the area under F(x) on $a \le x \le b$.
$$A=\int_a^b F(x) dx$$​

This means he's finding the area between the x-axis and the curve F(x) that is probably assumed to be above the x-axis.
This area is further bounded by a vertical line on the left side and a vertical line on the right side.