Intergration calculating centroid of an area

Hedgecutter

I'm revising at the moment and I have hit a wall with one question. I have to find the centroid of the area of a curve Y=3x^2 the x axis and ordinates x=0,y=2

looking over my notes and I dont understand how we found y bar.

I don't understand how we went from 3x^2 to 9x^4,

if we integrate it should be 3x^3/3?

I have attached my notes for y bar

any help would be appreciated

TheBody

Hedgecutter wrote: »

I'm revising at the moment and I have hit a wall with one question. I have to find the centroid of the area of a curve Y=3x^2 the x axis and ordinates x=0,y=2

looking over my notes and I dont understand how we found y bar.

I don't understand how we went from 3x^2 to 9x^4,

if we integrate it should be 3x^3/3?

I have attached my notes for y bar

any help would be appreciated

Hi there.

On the top of the second line, you just forgot the square when y got repalced. It should read [latex]\frac{1}{2}\int_0^2(3x^2)^2dx [/latex]

Hedgecutter

TheBody wrote: »

Hi there.

On the top of the second line, you just forgot the square when y got repalced. It should read [latex]\frac{1}{2}\int_0^2(3x^2)^2dx [/latex]

Cheers. so its 3*3=9 and add the multiply the powers 2*2=4 9^4

TheBody

Hedgecutter wrote: »

Cheers. so its 3*3=9 and add the multiply the powers 2*2=4 to give 9^4 [latex]9x^4[/latex]

FYP a bit!!

Hedgecutter

TheBody wrote: »

FYP a bit!!

thanks

on another note how do you use superscript on boards?

TheBody

Hedgecutter wrote: »

thanks

on another note how do you use superscript on boards?

Glad I could help.

You need to know how to use latex.

There is a forum on boards about it:

http://www.boards.ie/vbulletin/forumdisplay.php?f=1286

Hedgecutter

Question I'm working on at the moment.

The curve Y=4t-t^2 cuts the t axis at t=0 and t=4. Calculate the centroid of the area bounded by the curve and the t axis.

Not sure what I did in steps 2 and 3 are correct. I integrated twice above the line and only once below.

also not sure about (4t-t^2)^2 being = 16t^2t^4 and I cant go any further with the problem

TheBody

I'm afraid you have made a number of mistakes.

To begin, this time you are dealing with a function of t. Therefore, we have [latex]\overline{x}=\frac{\int_0^4(t)(4t-t^2)dt}{\int_0^44t-t^2dt}[/latex]. Notice that I don't have a mixture of x and t.
In general, you NEVER want a mixture of variables in an integral.

There are a number of errors in the [latex]\overline{y}[/latex] bit too.
To get you going, you should have:

[latex]\overline{y}=\frac{\frac{1}{2}\int_0^4(4t-t^2)^2dt}{\int_0^44t-t^2dt}=\frac{\frac{1}{2}\int_0^4t^4-8t^3+16t^2dt}{\int_0^44t-t^2dt}[/latex].

See if you can finish it off from there.

Hedgecutter

TheBody wrote: »

I'm afraid you have made a number of mistakes.

To begin, this time you are dealing with a function of t. Therefore, we have [latex]\overline{x}=\frac{\int_0^4(t)(4t-t^2)dt}{\int_0^44t-t^2dt}[/latex]. Notice that I don't have a mixture of x and t.
In general, you NEVER want a mixture of variables in an integral.

There are a number of errors in the [latex]\overline{y}[/latex] bit too.
To get you going, you should have:

[latex]\overline{y}=\frac{\frac{1}{2}\int_0^4(4t-t^2)^2dt}{\int_0^44t-t^2dt}=\frac{\frac{1}{2}\int_0^4t^4-8t^3+16t^2dt}{\int_0^44t-t^2dt}[/latex].

See if you can finish it off from there.

Ya I have a bad habit of mixing those up.

Could you write out long hand how you worked this out please.

[latex]\overline{y}=\frac{\frac{1}{2}\int_0^4(4t-t^2)^2dt}{\int_0^44t-t^2dt}=\frac{\frac{1}{2}\int_0^4t^4-8t^3+16t^2dt}{\int_0^44t-t^2dt}[/latex].

TheBody

Hedgecutter wrote: »

Ya I have a bad habit of mixing those up.

Could you write out long hand how you worked this out please.

[latex]\overline{y}=\frac{\frac{1}{2}\int_0^4(4t-t^2)^2dt}{\int_0^44t-t^2dt}=\frac{\frac{1}{2}\int_0^4t^4-8t^3+16t^2dt}{\int_0^44t-t^2dt}[/latex].

What you missed was that [latex](4t-t^2)^2=(4t-t^2)(4t-t^2)=t^4-8t^3+16t^2[/latex].

Notice that:

[latex](4t-t^2)^2\not=16t^2-t^4[/latex]

Hedgecutter

TheBody wrote: »

What you missed was that [latex](4t-t^2)^2=(4t-t^2)(4t-t^2)=t^4-8t^3+16t^2[/latex].

Notice that:

[latex](4t-t^2)^2\not=16t^2-t^4[/latex]

Basic stuff really. Once I get a few more under my belt I should be fine. thanks again, ill post the finished solution tomorrow.

TheBody

Hedgecutter wrote: »

Basic stuff really. Once I get a few more under my belt I should be fine. thanks again, ill post the finished solution tomorrow.

It'll all come to you with time. Keep your head up and you will do fine. What course are you doing? (if you don't mind me asking?)

Hedgecutter

TheBody wrote: »

It'll all come to you with time. Keep your head up and you will do fine. What course are you doing? (if you don't mind me asking?)

Mechanical engineering. Four modules this year. Advanced maths is one.

Hedgecutter

TheBody wrote: »

What you missed was that [latex](4t-t^2)^2=(4t-t^2)(4t-t^2)=t^4-8t^3+16t^2[/latex].

Notice that:

[latex](4t-t^2)^2\not=16t^2-t^4[/latex]

Question when I worked out this I got 16t^2-8t^3+t^4

does this make a difference?

TheBody

Hedgecutter wrote: »

Question when I worked out this I got 16t^2-8t^3+t^4

does this make a difference?

You got the same thing as me, only mine is in a different order. Well done!

Hedgecutter

TheBody wrote: »

You got the same thing as me, only mine is in a different order. Well done!

Do I need to do something else with it before I integrate?

TheBody

Hedgecutter wrote: »

Do I need to do something else with it before I integrate?

No, you are ready to integrate now.

Hedgecutter

not sure if this is right, answer looks a bit mad compared to X bar

TheBody

Sorry for the delay in getting back to you. You made a small mistake at the very beginning. You have an [latex]-18t^3[/latex] term when is should be an
[latex]-8t^3[/latex] term. You had it correct earlier; I guess you just wrote it down wrong.

Other than that I think it looks ok. Just note the once yu do the integrating bit you should drop the integral symbol and the dt. i.e from the second line onwards.

Almost there!!

P.S. I got [latex]\overline{y}=\frac{8}{5}[/latex] as the final answer.

Hedgecutter

TheBody wrote: »

Sorry for the delay in getting back to you. You made a small mistake at the very beginning. You have an [latex]-18t^3[/latex] term when is should be an
[latex]-8t^3[/latex] term. You had it correct earlier; I guess you just wrote it down wrong.

Other than that I think it looks ok. Just note the once yu do the integrating bit you should drop the integral symbol and the dt. i.e from the second line onwards.

Almost there!!

P.S. I got [latex]\overline{y}=\frac{8}{5}[/latex] as the final answer.

Ya that's what I get for trying to do it with kids around. I'll have a go again in the morning.
Part "b" looks interesting indefinite integral. Factorising could catch me again.

Hedgecutter

Hedgecutter wrote: »

Ya that's what I get for trying to do it with kids around. I'll have a go again in the morning.
Part "b" looks interesting indefinite integral. Factorising could catch me again.

Ya 8/5.

TheBody

Hedgecutter wrote: »

Ya 8/5.

Well done! Have you exams in January?

Hedgecutter

(B) next one I have. We usually factorise what's under the line but not sure how to factorise 1+x^2

Hedgecutter

TheBody wrote: »

Well done! Have you exams in January?

No June. Metrics, complex numbers, differentiation, intergration covered. Calculus by far the hardest. Other two handy enough. Starting laplace next week😥
Don't want to fall behind with calculus so trying to build up a stock of my own examples for revision in May.

TheBody

Hedgecutter wrote: »

(B) next one I have. We usually factorise what's under the line but not sure how to factorise 1+x^2

You can't factorise that. You don't need to anyway. This is just regular integration using substitution. Let [latex]u=1+x^2[/latex]. Give it a go and see how you get on.

Hedgecutter

TheBody wrote: »

You can't factorise that. You don't need to anyway. This is just regular integration using substitution. Let u=1+x^2. Give it a go and see how you get on.

Lovely I was making it hard for myself.

TheBody

Hedgecutter wrote: »

Lovely I was making it hard for myself.

When you do enough of them, you can spot the type of integral much faster.

Hedgecutter

Gave intergration by substitution a go. Think I'm ok at the start but not sure about the last line.

TheBody

The idea here is to turn an integration of [latex]x[/latex] into an integration depending totally on [latex]u[/latex]. As I said before, you never want to have a mixture of variables.

To get you started, we want:

[latex]\int\frac{3x}{1+x^2}dx=3\int\frac{x}{1+x^2}dx[/latex].

Remember we are trying to exchange all the x stuff. We already know that [latex]u=1+x^2[/latex], so we need to replace the [latex]xdx[/latex] bit.

As you did, [latex]\frac{du}{dx}=2x[/latex], therefore

[latex]xdx}=\frac{du}{2}[/latex]

We can now replace all the x stuff to get:

[latex]3\int\frac{1}{u}\frac{du}{2}=\frac{3}{2}\int\frac{1}{u}du[/latex].

Can you finish it from there?

Hedgecutter

TheBody wrote: »

The idea here is to turn an integration of [latex]x[/latex] into an integration depending totally on [latex]u[/latex]. As I said before, you never want to have a mixture of variables.

To get you started, we want:

[latex]\int\frac{3x}{1+x^2}dx=3\int\frac{x}{1+x^2}dx[/latex].

Remember we are trying to exchange all the x stuff. We already know that [latex]u=1+x^2[/latex], so we need to replace the [latex]xdx[/latex] bit.

As you did, [latex]\frac{du}{dx}=2x[/latex], therefore

[latex]xdx}=\frac{du}{2}[/latex]

We can now replace all the x stuff to get:

[latex]3\int\frac{1}{u}\frac{du}{2}=\frac{3}{2}\int\frac{1}{u}du[/latex].

Can you finish it from there?

[latex]xdx}=\frac{du}{2}[/latex] You cross multiply to get rid of the X from the 2

Hedgecutter

=3/2 LN +c

=3/2 LN [1+x^2]+c

Brackets are incorrect but all I can find.

TheBody

Hedgecutter wrote: »

=3/2 LN +c

=3/2 LN [1+x^2]+c

Brackets are incorrect but all I can find.

Well done. That's it all done.

Hedgecutter

Lovely. Thanks again for your help.

I have another turning point question think I might be close.

calculate the turning points and max and min value on the curve

Y=x^3-3x^3-9x+10.

Hedgecutter

using integration find the area between curves y=x2+1 and y=7-x

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
(x-3)(x+2)
x=-3 x=2

wondering if im on the right track

TheBody

I'm out on the sauce for New Year's Eve but I will reply to your posts tomorrow if nobody gets there before me. Happy new year!

Hedgecutter

TheBody wrote: »

I'm out on the sauce for New Year's Eve but I will reply to your posts tomorrow if nobody gets there before me. Happy new year!

Enjoy, cracking open a can here myself Happy New Year

TheBody

Hedgecutter wrote: »

Lovely. Thanks again for your help.

I have another turning point question think I might be close.

calculate the turning points and max and min value on the curve

Y=x^3-3x^3-9x+10.

Everything looks good in that one. The ony comment I'd make is that after testing the points in the second derivative to determine which point is the local max and min, you should write down which point is which. i.e say that (-1,15) is the local max and (3,-17) is the local min.

TheBody

Hedgecutter wrote: »

using integration find the area between curves y=x2+1 and y=7-x

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
(x-3)(x+2)
x=-3 x=2

wondering if im on the right track

That is almost correct. You made a small error with the signs of the factors. It should be [latex](x+3)(x-2)=0[/latex]. Giving you that x=-3 or x=2.

You have found the x coordinates where the two curves meet.

When finding the area betwen two curves, it's really useful to draw a rough sketch of the curves first. You need to know which curve is "on top" and which curve is on "the bottom".

Keep going and see how you get on.

Hedgecutter

TheBody wrote: »

This is fine so far. You have found the x coordinates where the two curves meet.

When finding the area betwen two curves, it's really useful to draw a rough sketch of the curves first. You need to know which curve is "on top" and which curve is on "the bottom".

Keep going and see how you get on.

Should I have x=0

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
x(x-3)(x+2)
x=0,x=-3, x=2







Looking at my notes I noticed that we sketched the curved to see which was positive and which was negative.



I'm I right in saying I need y coordinates?

X=1
Y=X^2+X-6
Y=(1)^2+(1)-6
Y=1+1-6
Y=-4

(1,-4)

I'm a bit confused why we need to know this

TheBody

I have just changed post #38. I missed a small mistake.

X=0 is NOT a root of the quadratic equation. What you had done the first time (apart from the small error pointed out in post #38 was correct.

With regards finding the y values, the ony use they serve is in helping you draw a more accurate sketch. They won't play any part in the actual integration.

Also in the post above, you ran x=1 through the function. I don't know why you did this. Run the x values found earlier through one of the functions. It doesn't matter which one you use.

Hedgecutter

TheBody wrote: »

I have just changed post #38. I missed a small mistake.

X=0 is NOT a root of the quadratic equation. What you had done the first time (apart from the small error pointed out in post #38 was correct.

With regards finding the y values, the ony use they serve is in helping you draw a more accurate sketch. They won't play any part in the actual integration.

Also in the post above, you ran x=1 through the function. I don't know why you did this. Run the x values found earlier through one of the functions. It doesn't matter which one you use.

x2 +1=7-x
x2+1-7+x=0
x2-6+x=0
X2+x-6=0
(x-3)(x+2)
x-3=0,x=3
x+2=0,x=-2
x=3, x=-2



On the formula I need to plug in the coordinates,

What i'm not sure of is where to get the other two coordinates.

Ill use two parts of the area formula, on that formula I have a,b,b,c where I plug in the coordinates are the other two coordinates 0?

TheBody

Hedgecutter wrote: »

On the formula I need to plug in the coordinates,

What i'm not sure of is where to get the other two coordinates.

Ill use two parts of the area formula, on that formula I have a,b,b,c where I plug in the coordinates are the other two coordinates 0?

I'm not sure what you mean by this.

What you need to do next is draw a sketch of curves. Then to find the area you work out:

[latex]\int_{-3}^2[/latex](upper curve)-(lower curve) dx

Hedgecutter

Sketched the graph and nothing below the line

Still confused to the importance of knowing this.

TheBody

Hedgecutter wrote: »

Sketched the graph and nothing below the line

Still confused to the importance of knowing this.

Sketch looks ok. You need to do this to find out which curve was "on top". Now you know the [latex]7-x[/latex] is on top and the [latex]x^2+1[/latex] is on the bottom (at least over the interval from -3 to 2).

At any rate now you find the area by working out:

[latex]\int_{-3}^2(7-x)-(x^2+1)dx[/latex]

Hedgecutter

TheBody wrote: »

Sketch looks ok. You need to do this to find out which curve was "on top". Now you know the [latex]7-x[/latex] is on top and the [latex]x^2+1[/latex] is on the bottom (at least over the interval from -3 to 2).

At any rate now you find the area by working out:

[latex]\int_{-3}^2(7-x)-(x^2+1)dx[/latex]

so because 7-x is on top it's plugged in to the formula first?

TheBody

Hedgecutter wrote: »

so because 7-x is on top it's plugged in to the formula first?

Yes, that's exactly it. The best way to see that is with a sketch.

Hedgecutter

TheBody wrote: »

Yes, that's exactly it. The best way to see that is with a sketch.

how do you know where the coordinates go a=-3 b=2 is it because -3 is first on the table working out y+7-x so -3=a?

Hedgecutter

Had a go at finishing

Hedgecutter

Had a go at finishing it

TheBody

Very close!! You made a small mistake near the beginning when you were removing the brackets. You should have,

[latex]\int_{-3}^2(7-x)-(x^2+1) dx=\int_{-3}^2 7-x-x^2-1 dx=\int_{-3}^2-x^2-x+6 dx[/latex].

Other than that your method was correct. Also, you should have equal signs as you progress from step to step.

TheBody

Hedgecutter wrote: »

how do you know where the coordinates go a=-3 b=2 is it because -3 is first on the table working out y+7-x so -3=a?

If you look at the sketch you made of the region, the most left x value is -3 and it runs up as far as x=2. Thats why I used those values.

Not sure if that makes sense??