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# Integration question watch

1. Stuck on part d) of the question below.

I found the area of the triangle which has PR as its hypotenuse (area of 5) and then subtracted the area of the little corner of the ellipse by integrating using the parametric equations.

Answer I got is 2.5pi but book has 10-2.5pi.

Not sure where I went wrong.

Could somebody check on the answer or confirm my method... If I've gone wrong I shall post further working. Thankyou.
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2. Looks like the solution has missed of the minus sign of the integral that comes from the .
3. (Original post by stanante)
Looks like the solution has missed of the minus sign of the integral that comes from the .
Oh no, I realised if I swap my limits around then I get the correct answer -why would I need to do it this way though?
4. It's been a long time since i've done this. We're integrating over the interval

So given that then for and for

So the integral becomes

Makes sense?
5. (Original post by stanante)
It's been a long time since i've done this. We're integrating over the interval

So given that then for and for

So the integral becomes

Makes sense?
Oh right that makes sense now - however is there a way or knowing how to arrange the limits in terms of the parameter without having to consider the x-values?
6. I don't immediately see how. Perhaps someone else can comment?
7. Anyone?
8. (Original post by fayled)
Anyone?
The method suggested by stanante is the only one I'm aware of.
9. (Original post by stanante)
I don't immediately see how. Perhaps someone else can comment?
Well the point is that when do a conventional "area" integral you are always integrating from left to right in the "x sense" so you do need to consider the x values. If you are integrating with respect to a parameter (as here), you need to establish which value of the parameter gives the leftmost x coordinate (this is the lower limit of integration), and which value gives the rightmost x coordinate (this is the upper limit of integration).

As a check, if the section of curve you're integrating lies above the x-axis then you would expect to get a positive value for the integral. If you come out with a negative answer, that's a sign that you've got your limits the wrong way round!
10. (Original post by davros)
Well the point is that when do a conventional "area" integral you are always integrating from left to right in the "x sense" so you do need to consider the x values. If you are integrating with respect to a parameter (as here), you need to establish which value of the parameter gives the leftmost x coordinate (this is the lower limit of integration), and which value gives the rightmost x coordinate (this is the upper limit of integration).

As a check, if the section of curve you're integrating lies above the x-axis then you would expect to get a positive value for the integral. If you come out with a negative answer, that's a sign that you've got your limits the wrong way round!
How does it work for say a spherical polar coordinate system. If you're integrating an area or volume do you still have to consider the cartesian values and make a change of variable?
11. (Original post by stanante)
How does it work for say a spherical polar coordinate system. If you're integrating an area or volume do you still have to consider the cartesian values and make a change of variable?
No, but you need to consider the "sense" in which you integrate with respect to a coordinate system - conventionally, a radius variable r goes from 0 to R for a finite volume (not the other way round), and an angular variable goes from 0 to 2pi not 2pi to 0.

(I can't say I actually remember a question that does spherical polars AND a parameterization of a curve in terms of a single parameter t.)
12. Could somebody take a look at part b) of this question please...

I'm confused because it seems inconsistent with the previous question I asked at the start of this thread (look at the response in post 4).

The limits of this integral in part b) are from pi/2 to 0 in terms of t but if you look at this in terms of x values the integral is from 0 to 3 (0 at top, 3 at bottom by integral sign) which seems to be the wrong way around.

I would have thought you would integrate from 0 to pi/2 thus giving an integral from 3 to 0 in terms of x.

However evaluating the integral in the question gives a +ve area so must be right, but why?

Maybe it is something to do with it being a closed loop? Thankyou.
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13. Isn't it just that the parameters are "the wrong way around", so A is a negative constant?

EDIT: Wait, you get a +ve area?
EDIT2: Are you sure sin2tsint is the function you're thinking of? Perhaps this one already goes from left to right as t increases.
14. (Original post by fayled)
Could somebody take a look at part b) of this question please...

I'm confused because it seems inconsistent with the previous question I asked at the start of this thread (look at the response in post 4).

The limits of this integral in part b) are from pi/2 to 0 in terms of t but if you look at this in terms of x values the integral is from 0 to 3 (0 at top, 3 at bottom by integral sign) which seems to be the wrong way around.

I would have thought you would integrate from 0 to pi/2 thus giving an integral from 3 to 0 in terms of x.

However evaluating the integral in the question gives a +ve area so must be right, but why?

Maybe it is something to do with it being a closed loop? Thankyou.
As I said, if you have y=f(x) you integrate from left to right in the x-sense to get the area. In (x,y) coordinates you are going from the point (0,0) to the point (3,0). This corresponds to going from t=pi/2 to t=0.

So:

15. (Original post by davros)
As I said, if you have y=f(x) you integrate from left to right in the x-sense to get the area. In (x,y) coordinates you are going from the point (0,0) to the point (3,0). This corresponds to going from t=pi/2 to t=0.

So:

Oh so I can remove a negative sign and compensate for it by swapping the limits? Thanks.

By the way, should I be reading your first integral as the integral from zero to three of y with respect to x (as I was previously saying from three to zero)...
16. (Original post by fayled)
Oh so I can remove a negative sign and compensate for it by swapping the limits? Thanks.

By the way, should I be reading your first integral as the integral from zero to three of y with respect to x (as I was previously saying from three to zero)...
Yes and yes!

It helps to play around with some simple integrals that you can do directly - e.g. of x, x^2 etc. Think what answer you normally get when you integrate "left to right", what you do with the limits when you plug them in at the end, and what you get if you put the limits in the other way round. This helps you get used to manipulating integrals and interpreting them.

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