I don’t really know where to start with this one – Though I can do the first squiggly line that has a zero at the bottom and one at the top, and the second squiggly line thing that has pi-on-three on the bottom and pi-on-six at the top.
The bit in between that and the answer is what I’m stuck on . Any suggestions please?

Spiny Norman said:

Any suggestions please?

I’m just checking in and will be out the rest of the day – moving house – but maybe, just maybe, you could show the actual equations.

sibeen said:

Spiny Norman said:

Any suggestions please?

I’m just checking in and will be out the rest of the day – moving house – but maybe, just maybe, you could show the actual equations.

That’s all there is.

Spiny Norman said:

I don’t really know where to start with this one – Though I can do the first squiggly line that has a zero at the bottom and one at the top, and the second squiggly line thing that has pi-on-three on the bottom and pi-on-six at the top.
The bit in between that and the answer is what I’m stuck on . Any suggestions please?

x = r.Cos(theta) and y =r.Sin(theta)

so xy = r^2.(Cos(theta).Sin(theta))

and integral dr from 0 to 1 = r^3/3.(Cos(theta).Sin(theta))

So you just need to integrate that with respect to theta.

Spiny Norman said:

sibeen said:

Spiny Norman said:

Any suggestions please?

I’m just checking in and will be out the rest of the day – moving house – but maybe, just maybe, you could show the actual equations.

That’s all there is.

All I can see is:

The Rev Dodgson said:

x = r.Cos(theta) and y =r.Sin(theta)

so xy = r^2.(Cos(theta).Sin(theta))

and integral dr from 0 to 1 = r^3/3.(Cos(theta).Sin(theta))

So you just need to integrate that with respect to theta.

So like this?

sibeen said:

Spiny Norman said:

sibeen said:

I’m just checking in and will be out the rest of the day – moving house – but maybe, just maybe, you could show the actual equations.

That’s all there is.

All I can see is:

Yep that’s right.xxxxxxxxxxxx

Dunno where all the x’s came from sorry.

Spiny Norman said:

The Rev Dodgson said:

x = r.Cos(theta) and y =r.Sin(theta)

so xy = r^2.(Cos(theta).Sin(theta))

and integral dr from 0 to 1 = r^3/3.(Cos(theta).Sin(theta))

So you just need to integrate that with respect to theta.

So like this?

Looks good to me.

But check your result with a numerical integration :)

Spiny Norman said:

I don’t really know where to start with this one – Though I can do the first squiggly line that has a zero at the bottom and one at the top, and the second squiggly line thing that has pi-on-three on the bottom and pi-on-six at the top.
The bit in between that and the answer is what I’m stuck on . Any suggestions please?

Just noticed:

How can Theta be greater than Pi/3 but less than Pi/6?

The Rev Dodgson said:

Spiny Norman said:

I don’t really know where to start with this one – Though I can do the first squiggly line that has a zero at the bottom and one at the top, and the second squiggly line thing that has pi-on-three on the bottom and pi-on-six at the top.
The bit in between that and the answer is what I’m stuck on . Any suggestions please?

Just noticed:

How can Theta be greater than Pi/3 but less than Pi/6?

I didn’t write it, maybe the arc goes around the long way.

Spiny Norman said:

The Rev Dodgson said:

Spiny Norman said:

I don’t really know where to start with this one – Though I can do the first squiggly line that has a zero at the bottom and one at the top, and the second squiggly line thing that has pi-on-three on the bottom and pi-on-six at the top.
The bit in between that and the answer is what I’m stuck on . Any suggestions please?

Just noticed:

How can Theta be greater than Pi/3 but less than Pi/6?

I didn’t write it, maybe the arc goes around the long way.

If it’s going round the long way wouldn’t it be pi/3 < Theta < (2pi + pi/6)?

Is there someone you can check with?

Has it been mentioned that dx * dy = r * dr * dθ ?

Don’t forget that extra factor of r.

ie.

∫ dy ∫ ρ dx = ∫ dx ∫ ρ dy = ∫ dθ ∫ ρ r dr = ∫ r dr ∫ ρ dθ

In this case you can integrate in either r first or θ first.

Ahh, I see where I was going wrong here….I was being stupid :)

In my defence I am reasonably unintelligent and haven’t done double integrals in many a year.

Spiny Norman said:

The Rev Dodgson said:

x = r.Cos(theta) and y =r.Sin(theta)

so xy = r^2.(Cos(theta).Sin(theta))

and integral dr from 0 to 1 = r^3/3.(Cos(theta).Sin(theta))

So you just need to integrate that with respect to theta.

So like this?

No, you’ve got the order of integration backwards (the order itself is not important (usually), but the application of limits is.) You’ve integrated over r, but applied the limits of dθ; the numbers on the RHS square brackets should be 0 (lower) and 1 (upper), and the π/3 and π/6 values should still be on the integral symbol.

btm said:

Spiny Norman said:

The Rev Dodgson said:

x = r.Cos(theta) and y =r.Sin(theta)

so xy = r^2.(Cos(theta).Sin(theta))

and integral dr from 0 to 1 = r^3/3.(Cos(theta).Sin(theta))

So you just need to integrate that with respect to theta.

So like this?

No, you’ve got the order of integration backwards (the order itself is not important (usually), but the application of limits is.) You’ve integrated over r, but applied the limits of dθ; the numbers on the RHS square brackets should be 0 (lower) and 1 (upper), and the π/3 and π/6 values should still be on the integral symbol.

oops.

Well spotted.

Order of integration doesn’t matter here.

What matters is that you’re integrating r * rho in polar coordinates.

So that’s an integral of r^3 cos sin not r^2 cos sin.

All about integrating Cos(theta)Sin(theta):

https://mindyourdecisions.com/blog/2015/08/09/the-perplexing-integral-of-sin-xcos-x-sunday-puzzle/

(a bit long, but it does give three equivalent solutions).

Just got back.
Have read the thread.
Can someone please post how the working goes, then the answer?

Spiny Norman said:

Just got back.
Have read the thread.
Can someone please post how the working goes, then the answer?

Spiny Norman said:

Just got back.
Have read the thread.
Can someone please post how the working goes, then the answer?

The integral was as your last post, but with the limits swapped, as pointed out by btm.

The integral of sin(theta)cos(theta) is in the link I posted. Just use the first one they give if you don’t want to read the whole thing.

That should give you enough to get the answer.

I haven’t read thread so sorry if already done.

BTW it is a bit weird that the end theta is lower than the start theta but whatevs. This means the answer will be negative.

0 <= r < = 1
pi/3 <= theta <= pi/6
f(x,y) = xy = r^2 cos(theta)*sin(theta)

so we want

int{theta = from pi/3 to pi/6, int{ r = from 0 to 1, r^2 cos(theta)*sin(theta)}}

if we integrate by r first:

int{theta = from pi/3 to pi/6 }

int{theta = from pi/3 to pi/6, (1/3 cos(theta)*sin(theta)} (by substituting and subtracting)

To integrate cos(theta)*sin(theta) you could use integration by parts or any number of other ways but anyway the integral of cos(theta)*sin(theta) is 1/2 (sin theta)^2

1/6 (0.25 – 0.75) (sub and sub)
= -1/12

I hope it’s right!

dv said:

I haven’t read thread so sorry if already done.

BTW it is a bit weird that the end theta is lower than the start theta but whatevs. This means the answer will be negative.

0 <= r < = 1
pi/3 <= theta <= pi/6
f(x,y) = xy = r^2 cos(theta)*sin(theta)

so we want

int{theta = from pi/3 to pi/6, int{ r = from 0 to 1, r^2 cos(theta)*sin(theta)}}

if we integrate by r first:

int{theta = from pi/3 to pi/6 }

int{theta = from pi/3 to pi/6, (1/3 cos(theta)*sin(theta)} (by substituting and subtracting)

To integrate cos(theta)*sin(theta) you could use integration by parts or any number of other ways but anyway the integral of cos(theta)*sin(theta) is 1/2 (sin theta)^2

1/6 (0.25 – 0.75) (sub and sub)
= -1/12

I hope it’s right!

I have no idea. :( Thanks anyway.

The Rev Dodgson said:

The integral was as your last post, but with the limits swapped, as pointed out by btm.

The integral of sin(theta)cos(theta) is in the link I posted. Just use the first one they give if you don’t want to read the whole thing.

That should give you enough to get the answer.

Okay I swapped them over and fed it into here

Still confusing as hell. Maybe I’ll try swapping over the integrals and d thingies and see how that goes.

We can make it a bit simpler just by separating the two variables, which we can do because the limits don’t involve each other and the function is just a product of two variables.

dv said:

We can make it a bit simpler just by separating the two variables, which we can do because the limits don’t involve each other and the function is just a product of two variables.

I have come to a better solution – Just tell the tutor that I don’t know how to do it. I know you guys find this stuff fairly easy but I’m sorry I just don’t. It’s been about 35 years since I went to school and we never did any calculus then either, unlike the senior school kids these days.

Thanks for trying anyway.

That’s what I get

dv said:

That’s what I get

Yeah the site I linked to gets the same answer, ta.

dv said:

I haven’t read thread so sorry if already done.

BTW it is a bit weird that the end theta is lower than the start theta but whatevs. This means the answer will be negative.

0 <= r < = 1
pi/3 <= theta <= pi/6
f(x,y) = xy = r^2 cos(theta)*sin(theta)

so we want

int{theta = from pi/3 to pi/6, int{ r = from 0 to 1, r^2 cos(theta)*sin(theta)}}

if we integrate by r first:

int{theta = from pi/3 to pi/6 }

int{theta = from pi/3 to pi/6, (1/3 cos(theta)*sin(theta)} (by substituting and subtracting)

To integrate cos(theta)*sin(theta) you could use integration by parts or any number of other ways but anyway the integral of cos(theta)*sin(theta) is 1/2 (sin theta)^2

1/6 (0.25 – 0.75) (sub and sub)
= -1/12

I hope it’s right!

That’s what I got.

And here’s a little side bar on one way of working out the integral of sin theta cos theta.

The reason you might choose this method is that the integrand is a product of a function (sin theta) and its derivative (cos theta).

Spiny Norman said:

dv said:

We can make it a bit simpler just by separating the two variables, which we can do because the limits don’t involve each other and the function is just a product of two variables.

I have come to a better solution – Just tell the tutor that I don’t know how to do it.

Well that’s always an option.

Spiny Norman said:

I know you guys find this stuff fairly easy

I wouldn’t say that.

I just happen to have been working on some spreadsheet functions to do multiple linear integration, so it fits in with that.

The Rev Dodgson said:

Spiny Norman said:

I know you guys find this stuff fairly easy

I wouldn’t say that.

I just happen to have been working on some spreadsheet functions to do multiple linear integration, so it fits in with that.

I had to be given a major hint to have any idea as to what was being asked.

The Rev Dodgson said:

Spiny Norman said:

I know you guys find this stuff fairly easy

I wouldn’t say that.

I just happen to have been working on some spreadsheet functions to do multiple linear integration, so it fits in with that.

What I’ll probably end up doing is just not worry about this circular stuff and spend more time on the things that I can do.

Spiny Norman said:

The Rev Dodgson said:

Spiny Norman said:

I know you guys find this stuff fairly easy

I wouldn’t say that.

I just happen to have been working on some spreadsheet functions to do multiple linear integration, so it fits in with that.

What I’ll probably end up doing is just not worry about this circular stuff and spend more time on the things that I can do.

Do what you can in the Maths exam and hand it in with a folded Mawson inside.

Mollwoll said something and we missed it:

“What matters is that you’re integrating r * rho in polar coordinates.

So that’s an integral of r^3 cos sin not r^2 cos sin.”

I think the result should be -1/16, not -1/12.

The Rev Dodgson said:

Mollwoll said something and we missed it:

“What matters is that you’re integrating r * rho in polar coordinates.

So that’s an integral of r^3 cos sin not r^2 cos sin.”

I think the result should be -1/16, not -1/12.

I’m obviously missing something, why r³

sibeen said:

The Rev Dodgson said:

Mollwoll said something and we missed it:

“What matters is that you’re integrating r * rho in polar coordinates.

So that’s an integral of r^3 cos sin not r^2 cos sin.”

I think the result should be -1/16, not -1/12.

I’m obviously missing something, why r³

It would be good to see the full question as asked, rather than just the bounding params.

dv said:

sibeen said:

The Rev Dodgson said:

Mollwoll said something and we missed it:

“What matters is that you’re integrating r * rho in polar coordinates.

So that’s an integral of r^3 cos sin not r^2 cos sin.”

I think the result should be -1/16, not -1/12.

I’m obviously missing something, why r³

It would be good to see the full question as asked, rather than just the bounding params.

Specifically:

are we being asked to integrate this function with respect to theta and r between the specified limits

or

are we being asked to integrate this function over the area _indicated?

dv said:

dv said:

sibeen said:

I’m obviously missing something, why r³

It would be good to see the full question as asked, rather than just the bounding params.

Specifically:

are we being asked to integrate this function with respect to theta and r between the specified limits

or

are we being asked to integrate this function over the area _indicated?

Fair question.

SN doesn’t have any more info, but my interpretation would now be that we should be integrating xy over the area of the circular segment defined by pi/6, pi/3 and r= 1.

The Rev Dodgson said:

dv said:

dv said:

It would be good to see the full question as asked, rather than just the bounding params.

Specifically:

are we being asked to integrate this function with respect to theta and r between the specified limits

or

are we being asked to integrate this function over the area _indicated?

Fair question.

SN doesn’t have any more info, but my interpretation would now be that we should be integrating xy over the area of the circular segment defined by pi/6, pi/3 and r= 1.

I suppose that is most likely.

dv said:

The Rev Dodgson said:

dv said:

Specifically:

are we being asked to integrate this function with respect to theta and r between the specified limits

or

are we being asked to integrate this function over the area _indicated?

Fair question.

SN doesn’t have any more info, but my interpretation would now be that we should be integrating xy over the area of the circular segment defined by pi/6, pi/3 and r= 1.

I suppose that is most likely.

So why then the pi/3 to pi/6 kerfuffle?

sibeen said:

The Rev Dodgson said:

Mollwoll said something and we missed it:

“What matters is that you’re integrating r * rho in polar coordinates.

So that’s an integral of r^3 cos sin not r^2 cos sin.”

I think the result should be -1/16, not -1/12.

I’m obviously missing something, why r³

We are integrating a value with dimensions L^2 over an area, so the result must have dimensions of L^4.

sibeen said:

dv said:

The Rev Dodgson said:

Fair question.

SN doesn’t have any more info, but my interpretation would now be that we should be integrating xy over the area of the circular segment defined by pi/6, pi/3 and r= 1.

I suppose that is most likely.

So why then the pi/3 to pi/6 kerfuffle?

I think that’s either a mistake or a clever ploy to confuse the unwary.

sibeen said:

dv said:

The Rev Dodgson said:

Fair question.

SN doesn’t have any more info, but my interpretation would now be that we should be integrating xy over the area of the circular segment defined by pi/6, pi/3 and r= 1.

I suppose that is most likely.

So why then the pi/3 to pi/6 kerfuffle?

Well perhaps someone simply made a mistake.

But anyway, I concur with moll and Rev. IF indeed they are asking to integrate the function over the area, then it is 1/16. It they are asking to integrate the function, with respect to the stated variables, between the stated limits, then the answers is -1/12.

Got another session with the tutor tomorrow at 4pm. The question will be reviewed and I’ll let you all know some time after that.

Spiny Norman said:

Got another session with the tutor tomorrow at 4pm. The question will be reviewed and I’ll let you all know some time after that.

Ta.

Spiny Norman said:

Got another session with the tutor tomorrow at 4pm. The question will be reviewed and I’ll let you all know some time after that.

Well I hope he gets it right :)

I have now done a numerical double integration of xy over the limits:
x: 0 to 0.866025404
y: tan(pi/6)*x to tan(pi/3)*x if x < 0.5 else (1-x^2)^0.5

That defines a circular segment of radius 1 with centre at the origin and sides at pi/6 (30 degrees) and pi/3 (60 degrees).

and the result is
drumroll …….

0.0625 = 1/16

I have also learned how to write an if then else statement on a single line in Python, so it was all worthwhile.

SN, do you happen to remember the question being asked?

dv said:

SN, do you happen to remember the question being asked?

It’s in the first photo I posted, that’s all there is.

Spiny Norman said:

Got another session with the tutor tomorrow at 4pm. The question will be reviewed and I’ll let you all know some time after that.

drums fingers

sibeen said:

Spiny Norman said:

Got another session with the tutor tomorrow at 4pm. The question will be reviewed and I’ll let you all know some time after that.

drums fingers

How are your fingers now?

The Rev Dodgson said:

sibeen said:

Spiny Norman said:

Got another session with the tutor tomorrow at 4pm. The question will be reviewed and I’ll let you all know some time after that.

drums fingers

How are your fingers now?

Nubs.

sibeen said:

The Rev Dodgson said:

sibeen said:

drums fingers

How are your fingers now?

Nubs.

This is tiring.

sibeen said:

sibeen said:

The Rev Dodgson said:

How are your fingers now?

Nubs.

This is tiring.

Should use drumsticks.

Sorry I forgot to post that yesterday’s tutorial was cancelled, it’s now on at 4 pm today.

Now he tells me.

The mystery is solved – When I pointed it out, he said, “oh yeah I got those back to front”.
So there ya go.

Spiny Norman said:

The mystery is solved – When I pointed it out, he said, “oh yeah I got those back to front”.
So there ya go.

OK, but what’s the answer?

The Rev Dodgson said:

Spiny Norman said:

The mystery is solved – When I pointed it out, he said, “oh yeah I got those back to front”.
So there ya go.

OK, but what’s the answer?

Try turning the paper around?

The Rev Dodgson said:

Spiny Norman said:

The mystery is solved – When I pointed it out, he said, “oh yeah I got those back to front”.
So there ya go.

OK, but what’s the answer?

Bill, there was an issue as to whether the answer was 1/12 or 1/16, and it depended upon actual question.

sibeen said:

The Rev Dodgson said:

Spiny Norman said:

The mystery is solved – When I pointed it out, he said, “oh yeah I got those back to front”.
So there ya go.

OK, but what’s the answer?

Bill, there was an issue as to whether the answer was 1/12 or 1/16, and it depended upon actual question.

To wit:

If he was asking you to integrate the function “over the area”, the answer is 1/16. If he was asking you to integrate the function with respect to the given variables, within the given bounds, then the answer is 1/12. Or rather -1/12, if we include the typo.

Don’t know for sure sorry, we didn’t work on it after that. I’m guessing it was area that he was after as the other questions were along those lines.

Anyway, I now know more about assigning numbers to functions than I ever did. Thanks.

Spiny Norman said:

The mystery is solved – When I pointed it out, he said, “oh yeah I got those back to front”.
So there ya go.

That makes more sense.