Integration question

Looks like the solution has missed of the minus sign of the integral that comes from the

-5\sin \theta

.
This gives

10-\frac{5\pi }{2}

instead.
I got your answer of

\frac{5\pi }{2}

Reply 2

Original post by stanante

Looks like the solution has missed of the minus sign of the integral that comes from the

-5\sin \theta

.
This gives

10-\frac{5\pi }{2}

instead.
I got your answer of

\frac{5\pi }{2}

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?

Reply 3

It's been a long time since i've done this. We're integrating over the interval

\displaystyle \int_{\frac{5}{\sqrt{2}}}^{5}y\: dx

So given that

x=5\cos \theta

then for

x=5\; \Rightarrow \; \theta =0

and for

x=5/\sqrt{2}\; \Rightarrow \; \theta =\pi /4

So the integral becomes

\displaystyle -20\int_{\frac{\pi }{4}}^{0}\sin^{2}\theta \: d\theta

Makes sense?

Reply 4

Original post by stanante

It's been a long time since i've done this. We're integrating over the interval

\displaystyle \int_{\frac{5}{\sqrt{2}}}^{5}y\: dx

So given that

x=5\cos \theta

then for

x=5\; \Rightarrow \; \theta =0

and for

x=5/\sqrt{2}\; \Rightarrow \; \theta =\pi /4

So the integral becomes

\displaystyle -20\int_{\frac{\pi }{4}}^{0}\sin^{2}\theta \: d\theta

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?

Reply 5

I don't immediately see how. Perhaps someone else can comment?

Anyone?

Original post by fayled

Anyone?

The method suggested by stanante is the only one I'm aware of.

Reply 8

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!

Reply 9

Original post by davros

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?

Reply 10

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.)

Reply 11

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.

(edited 11 years ago)

C4.jpg90.6KB

Reply 12

aznkid66

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.

(edited 11 years ago)

Reply 13

Original post by fayled

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:

\displaystyle \int^{x=3}_{x=0} y dx = \int^{t=0}_{t=pi/2} y(t) (dx/dt) dt = \int^{t=0}_{t=pi/2} (9sin2t)(-3sint)dt = \int^{t=pi/2}_{t=0}27sin2tsint dt

Reply 14

Original post by davros

\displaystyle \int^{x=3}_{x=0} y dx = \int^{t=0}_{t=pi/2} y(t) (dx/dt) dt = \int^{t=0}_{t=pi/2} (9sin2t)(-3sint)dt = \int^{t=pi/2}_{t=0}27sin2tsint dt

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)...

(edited 11 years ago)

Reply 15