Finding Area using parametric equation

[songoku]

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

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

 

[docnet]

Hint: look for a point on the curve where $t=-1$ and find another point where $t=2$.

 

[songoku]

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

 

[pasmith]

This is an application of Green's theorem:
<br /> \iint_R \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\,dx\,dy = \oint_{\partial R} L\,dx + M\,dy where \partial R is the boundary of R traversed counter-clockwise. If you choose L and M such that \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1 then the line integral gives the area of R. Here taking M = 0 and L = -y gives the contribution from C as <br /> \int_C (-y)\,dx = -\int_2^{-1} (3 + 3t)\,2t\,dt = \int_{-1}^2 (3 + 3t)\,2t\,dt. There will also be a contribution from the line QP where (x,y) = (5,9) + (3, -4)t. The portion along the x-axis from the unlabelled point of intersection with C to the point Q does not contribute as y = 0 here.

 

[songoku]

This is an application of Green's theorem:
<br /> \iint_R \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\,dx\,dy = \oint_{\partial R} L\,dx + M\,dy where \partial R is the boundary of R traversed counter-clockwise. If you choose L and M such that \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1 then the line integral gives the area of R. Here taking M = 0 and L = -y gives the contribution from C as <br /> \int_C (-y)\,dx = -\int_2^{-1} (3 + 3t)\,2t\,dt = \int_{-1}^2 (3 + 3t)\,2t\,dt. There will also be a contribution from the line QP where (x,y) = (5,9) + (3, -4)t. The portion along the x-axis from the unlabelled point of intersection with C to the point Q does not contribute as y = 0 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

 

[Mark44]

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.

 

[docnet]

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 $\left\{-1<t<2\right\}$

 

[pasmith]

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 R with a straightforward integral of the form \int y\,dx because the curve C turns back on itself. You would have to split it into the area to the left of x = 2 where R is bounded by C for t \in [-1,0] and C for t \in [0,1], the area between x = 2 and x = 5 where R is bounded by the x-axis and C for t \in [1,2], and the triangle to the right of x = 5. You would end up with <br /> \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 $x = 5$}).

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

 

[Mark44]

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.

 

[songoku]

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 $t=2$

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 $t=0$?

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

You could do the integral as \int x\,dy between QP and C, but \int_{-1}^2 (3 - 3t)\,2t\,dt 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

 

[songoku]

I think I get it. Thank you very much docnet, pasmith, Mark44

 

[Mark44]

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.

 

[songoku]

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