Finding Area using parametric equation

In summary: So it is not possible to solve it without converting it to cartesian equation?No, it is not possible to solve it without converting it to cartesian equation. The reason is that the parametric equations do not have a simple representation in terms of x and y, so you cannot simply integrate over x and y. You need to convert the equations to a form where you can integrate in terms of x and y, which is the cartesian form.
  • #1
songoku
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Homework Statement
Given the equation of the graph in parametric form:
##x=t^2+1##
##y=3+3t##

Normal at point P (5, 9) cuts x-axis at Q. Find area R
Relevant Equations
Integration
1612338727075.png

I want to ask about the solution. The solution divides region R into two parts: curved part and triangle. The triangle is obtained by drawing line ##x=5##. Let say line ##x=5## cuts x-axis at point A so the triangle is PAQ

For the curved part:
$$\int_{-1}^{2} (3+3t) ~2t~ dt$$

My question:
Why the limit of integration for curved part is from ##t=-1## to ##t=2##? The value of ##x## at the leftmost part of the graph is ##x=1## so why the limit is not from ##t=0## to ##t=2##?

In my imagination, the curved area covered by the solution is only from ##x=2## to ##x=5## so there are region R from ##x=1## to ##x=2## that is not covered.

Thanks
 
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  • #2
Hint: look for a point on the curve where ##t=-1## and find another point where ##t=2##.
 
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  • #3
docnet said:
Hint: look for a point on the curve where ##t=-1## and find another point where ##t=2##.
For ##t=-1## , ##x=2## and ##y=0##. This is point where the curve intersects x-axis

For ##t=2##, ##x=5## and ##y=9##. This is point P

So it means that the integral does not cover the area from ##x=1## to ##x=2##?

Thanks
 
  • #4
This is an application of Green's theorem:
[tex]
\iint_R \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\,dx\,dy = \oint_{\partial R} L\,dx + M\,dy[/tex] where [itex]\partial R[/itex] is the boundary of [itex]R[/itex] traversed counter-clockwise. If you choose [itex]L[/itex] and [itex]M[/itex] such that [tex]\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1[/tex] then the line integral gives the area of [itex]R[/itex]. Here taking [itex]M = 0[/itex] and [itex]L = -y[/itex] gives the contribution from [itex]C[/itex] as [tex]
\int_C (-y)\,dx = -\int_2^{-1} (3 + 3t)\,2t\,dt = \int_{-1}^2 (3 + 3t)\,2t\,dt.[/tex] There will also be a contribution from the line [itex]QP[/itex] where [itex](x,y) = (5,9) + (3, -4)t[/itex]. The portion along the x-axis from the unlabelled point of intersection with [itex]C[/itex] to the point [itex]Q[/itex] does not contribute as [itex]y = 0[/itex] here.
 
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  • #5
pasmith said:
This is an application of Green's theorem:
[tex]
\iint_R \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\,dx\,dy = \oint_{\partial R} L\,dx + M\,dy[/tex] where [itex]\partial R[/itex] is the boundary of [itex]R[/itex] traversed counter-clockwise. If you choose [itex]L[/itex] and [itex]M[/itex] such that [tex]\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1[/tex] then the line integral gives the area of [itex]R[/itex]. Here taking [itex]M = 0[/itex] and [itex]L = -y[/itex] gives the contribution from [itex]C[/itex] as [tex]
\int_C (-y)\,dx = -\int_2^{-1} (3 + 3t)\,2t\,dt = \int_{-1}^2 (3 + 3t)\,2t\,dt.[/tex] There will also be a contribution from the line [itex]QP[/itex] where [itex](x,y) = (5,9) + (3, -4)t[/itex]. The portion along the x-axis from the unlabelled point of intersection with [itex]C[/itex] to the point [itex]Q[/itex] does not contribute as [itex]y = 0[/itex] here.
Can the solution be explained without involving Green's theorem? I have not learned that, what I have learned is only integration (by part, substitution, trigonometry) and parametric equation

Thanks
 
  • #6
songoku said:
Can the solution be explained without involving Green's theorem?
Yes.
What I would do is to convert the parametric equations into an equation with y in terms of x. From that equation you can find the derivative dy/dx, and evaluate it at the point (5, 9) to get the slope of the tangent line and the normal at this point.

With these things done, you will have equations for the curve and the line boundary, and can set up an integral that will give the area of the enclosed region.

As a hint, the curve is that of a parabola opening to the right.
 
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  • #7
My way is not the cleverest way to solve this problem, and it may not even be right. As Pasmith says Green's theorem would be the way to solve it.

but you can arrive at the integral
$$\int_{-1}^2 (3+3t)2t dt$$
by doing a simple change of variables. Normally you are computing ##\int_{x_o}^{x}ydx## for the area under a graph in ##R^2##

because
##x=t^2+1##
we know that
##dx=2tdt##

Then plug in ##y## and ##dx## to get the integral in terms of ##t## and ##dt##. Determine the domain of integration by looking for the endpoints of the curve ##y## that we are integrating, that is ##\left\{-1<t<2\right\}##
 
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  • #8
songoku said:
Can the solution be explained without involving Green's theorem? I have not learned that, what I have learned is only integration (by part, substitution, trigonometry) and parametric equation

Thanks

No. You cannot compute the area of [itex]R[/itex] with a straightforward integral of the form [itex]\int y\,dx[/itex] because the curve [itex]C[/itex] turns back on itself. You would have to split it into the area to the left of [itex]x = 2[/itex] where [itex]R[/itex] is bounded by [itex]C[/itex] for [itex]t \in [-1,0][/itex] and [itex]C[/itex] for [itex]t \in [0,1][/itex], the area between [itex]x = 2[/itex] and [itex]x = 5[/itex] where [itex]R[/itex] is bounded by the x-axis and [itex]C[/itex] for [itex]t \in [1,2][/itex], and the triangle to the right of [itex]x = 5[/itex]. You would end up with [tex]
\int_0^1 (3 - 3t)\,2t\,dt - \int_{-1}^0 (3 - 3t)\,2t\,dt + \int_1^2 (3 - 3t)\,2t\,dt + (\mbox{area of triangle to right of $x = 5$}).[/tex]

You could do the integral as [itex]\int x\,dy[/itex] between QP and C, but [itex]\int_{-1}^2 (3 - 3t)\,2t\,dt[/itex] is not of that form.
 
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  • #9
pasmith said:
You cannot compute the area of R with a straightforward integral of the form ∫ydx
But you could compute the area of the region with an integral of the form ##\int x_{right} - x_{left} dy##, where the limits of integration are y = 0 and y = 5, and the x values in the integrand are those on the line and on the parabola (both written as functions of y). This integral would use horizontal strips. If you use vertical strips, you would need two separate integrals, as @pasmith wrote.
 
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  • #10
Mark44 said:
Yes.
What I would do is to convert the parametric equations into an equation with y in terms of x. From that equation you can find the derivative dy/dx, and evaluate it at the point (5, 9) to get the slope of the tangent line and the normal at this point.

With these things done, you will have equations for the curve and the line boundary, and can set up an integral that will give the area of the enclosed region.

As a hint, the curve is that of a parabola opening to the right.

I did the integration in cartesian equation as you suggested and got the same result as the solution but I still don't understand why the curved part in the solution is integrated from ##t=-1## to ##t=2## instead of from ##t=0## to ##t=2##

docnet said:
My way is not the cleverest way to solve this problem, and it may not even be right. As Pasmith says Green's theorem would be the way to solve it.

but you can arrive at the integral
$$\int_{-1}^2 (3+3t)2t dt$$
by doing a simple change of variables. Normally you are computing ##\int_{x_o}^{x}ydx## for the area under a graph in ##R^2##

because
##x=t^2+1##
we know that
##dx=2tdt##

Then plug in ##y## and ##dx## to get the integral in terms of ##t## and ##dt##. Determine the domain of integration by looking for the endpoints of the curve ##y## that we are integrating, that is ##\left\{-1<t<2\right\}##

Same question, why the limit starts from ##t=-1## instead of ##t=0##?

pasmith said:
No. You cannot compute the area of [itex]R[/itex] with a straightforward integral of the form [itex]\int y\,dx[/itex] because the curve [itex]C[/itex] turns back on itself. You would have to split it into the area to the left of [itex]x = 2[/itex] where [itex]R[/itex] is bounded by [itex]C[/itex] for [itex]t \in [-1,0][/itex] and [itex]C[/itex] for [itex]t \in [0,1][/itex], the area between [itex]x = 2[/itex] and [itex]x = 5[/itex] where [itex]R[/itex] is bounded by the x-axis and [itex]C[/itex] for [itex]t \in [1,2][/itex], and the triangle to the right of [itex]x = 5[/itex]. You would end up with [tex]
\int_0^1 (3 - 3t)\,2t\,dt - \int_{-1}^0 (3 - 3t)\,2t\,dt + \int_1^2 (3 - 3t)\,2t\,dt + (\mbox{area of triangle to right of $x = 5$}).[/tex]

You could do the integral as [itex]\int x\,dy[/itex] between QP and C, but [itex]\int_{-1}^2 (3 - 3t)\,2t\,dt[/itex] is not of that form.

Ah, I think I get it from your explanation. Curved part (from ##x=1## to ##x=5##) will be:

$$\int_0^1 (3 + 3t)\,2t\,dt - \int_{-1}^0 (3 + 3t)\,2t\,dt + \int_1^2 (3 + 3t)\,2t\,dt$$+
$$=\int_0^1 (3 + 3t)\,2t\,dt + \int_{0}^{-1} (3 + 3t)\,2t\,dt + \int_1^2 (3 + 3t)\,2t\,dt$$
$$=\int_{-1}^2 (3 + 3t)\,2t\,dt$$

I think my question has been resolved.

Edit: Oops, I just realized my working is wrong. I will try to do it first
 
Last edited:
  • #11
I think I get it. Thank you very much docnet, pasmith, Mark44
 
  • #12
songoku said:
I did the integration in cartesian equation as you suggested and got the same result as the solution but I still don't understand why the curved part in the solution is integrated from t=−1 to t=2 instead of from t=0 to
Because as t ranges from -1 to 2, the points (x, y) on the parabola are traced out from the x-intercept at (2, 0) up to point P at (5, 9).
If t = -1, x = 2, y = 0
If t = 0, x = 1, y = 3
If t = 2, x = 5, y = 9
For the intermediate values of t, we get all the points on that section of the parabola.
 
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  • #13
Mark44 said:
Because as t ranges from -1 to 2, the points (x, y) on the parabola are traced out from the x-intercept at (2, 0) up to point P at (5, 9).
If t = -1, x = 2, y = 0
If t = 0, x = 1, y = 3
If t = 2, x = 5, y = 9
For the intermediate values of t, we get all the points on that section of the parabola.
I understand. Thank you once again, Mark44
 

Related to Finding Area using parametric equation

1. What is the concept of "Finding Area using parametric equation"?

The concept of finding area using parametric equation is a method of calculating the area of a region using parametric equations. Parametric equations are a set of equations that express the coordinates of a point in terms of one or more parameters. These equations are often used to describe the motion of a point or object in a plane. By finding the area enclosed by the curve described by these equations, we can determine the total area of a region.

2. How do you find the area using parametric equations?

To find the area using parametric equations, we first need to determine the limits of integration for the parametric equations. This is done by finding the values of the parameters at which the curve begins and ends. Then, we can use the formula for finding the area under a curve, which is the integral of the function with respect to the variable of integration. In this case, the integral will be with respect to the parameter. Once the integral is evaluated, we will have the area of the region enclosed by the parametric equations.

3. What are the advantages of using parametric equations to find area?

Using parametric equations to find area has several advantages. One advantage is that it allows us to find the area of complex shapes that cannot be easily expressed using traditional equations. Additionally, parametric equations can be used to find the area of regions that are not easily described using Cartesian coordinates. This method also allows us to find the area of regions that are not necessarily bounded by straight lines, making it a versatile tool for finding area.

4. What are some common applications of finding area using parametric equations?

Finding area using parametric equations has many practical applications in fields such as physics, engineering, and mathematics. For example, it can be used to calculate the area of a region bounded by a parametric curve, such as the area under a projectile's path. It can also be used to find the area of a region enclosed by a parametric surface, which is useful in fields such as computer graphics and fluid dynamics.

5. Are there any limitations to using parametric equations to find area?

While parametric equations are a powerful tool for finding area, they do have some limitations. One limitation is that they are not always easy to work with, especially for complex shapes. Additionally, the parameterization of a curve or surface may not always be unique, which can lead to different results for the area calculation. It is important to carefully consider the parametric equations being used and their limitations when using this method to find area.

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