hmmms (2 Viewers)

[fullonoob]

evaluate integral between3 and -3 of y = root (9 -x^2)

 

[SeCKSiiMiNh]

change the limits to 3 and 0

then add a 2 next the root thing.

should work

oh , and let x = 3 sin theta

and the rest you should be able to do yourself

 

[fullonoob]

 

[fullonoob]

change the limits to 3 and 0

then add a 2 next the root thing.

should work

oh , and let x = 3 sin theta

and the rest you should be able to do yourself

working out please

 

[SeCKSiiMiNh]

yes i understand.

do what i said and see if you get the asnwer.

 

[adomad]

hmm.. well it is a semi circle with radius three... use the area of a circle formula

 

[Carrotsticks]

please be polite next time by stating that you have a problem, and then asking for help, rather than just chucking us the question.

Your question doesn't state whether you're finding the area with the X or the Y axis. But since the graph is clearly a semicircle on the X axis, it clearly must be with the X axis.

There are 3 methods:

1. Evaluate the integral normally with limits -3, 3. If you can't do this, you're in trouble.

2. Evaluate the integral with limits 0, 3 and then multiply the area by 2 (since its symmetrical about Y axis)

3. Realise that the shape is a semicircle with radius 3 units. Hence area is simply 9pi/2 units squared.

 

[fullonoob]

hmm.. well it is a semi circle with radius three... use the area of a circle formula

yes i know that but intergrate it
i've tried inverse chain rule and still failed miserably
gna take a break and see if i can get back to it

 

[life92]

With this type of integral, try to think of what the graph is.
It's a positive semi circle with centre (0,0) and radius 3.
This means that the integral is equal to the area of a semi circle radius 3.
Therefore, 9pi/2.


edit: beaten XD

 

[fullonoob]

please be polite next time by stating that you have a problem, and then asking for help, rather than just chucking us the question.

Your question doesn't state whether you're finding the area with the X or the Y axis. But since the graph is clearly a semicircle on the X axis, it clearly must be with the X axis.

There are 3 methods:

1. Evaluate the integral normally with limits -3, 3. If you can't do this, you're in trouble.

2. Evaluate the integral with limits 0, 3 and then multiply the area by 2 (since its symmetrical about Y axis)

3. Realise that the shape is a semicircle with radius 3 units. Hence area is simply 9pi/2 units squared.

I think you're confused with volume rotation. You can find area no matter what axis you are using, it is simply a matter of applying it.

 

[fullonoob]

ehh can anyone not do the cheapy method of area of circle?
and actually try to integrate it and find the area...will be much appreciated.

 

[life92]

If you want working for the integral.

It's easier to change the limits to 0 and 3, and multiply the answer by 2

Let x = 3sin theta
dx = 3cos theta d(theta)

So,
S (9-x^2)^(1/2) dx
= S (9-9sin^2 theta)^(1/2) 3 cos theta
= S [9(1-sin^2 theta)]^(1/2) 3 cos theta
= S 3 (cos^2 theta)^(1/2) 3 cos theta
= 9 S cos^2 theta

And you should be able to do it from here remembering that the limits are 0 and 3 to make it easier

 

[Carrotsticks]

I think you're confused with volume rotation. You can find area no matter what axis you are using, it is simply a matter of applying it.

Yeah i realised that right after I clicked 'post', but i dont know how to edit posts.

 

[xV1P3R]

Generally with a question like this, you are expected to recognise the graph and take the alternate path to attaining the answer. I don't think they would ask you to find the area through purely integration.

 

[life92]

Generally with a question like this, you are expected to recognise the graph and take the alternate path to attaining the answer. I don't think they would ask you to find the area through purely integration.

+1

I mean, if you can achieve the answer in like 2 lines, rather than ~6 wouldn't you rather just take the shorter path?
Unless the question specifically asks you to solve it by integration.

Where did you find this question btw?
I'm pretty sure that if they didn't give you the substitution it'd make it a 4U question.

 

[fullonoob]

thanks i was just curious how you do it 4u style. Just wondering now can you do the reverse chain rule with roots? or only with >1 indices

 

[life92]

You can do the reverse chain rule with any integral, as long as the derivative of the function is there.
But with this integral no.

If it was say, S x (9-x^2)^(1/2) dx instead, with the x at the front, then yes you could do the reverse chain rule, because the derivative of 9-x^2 is -2x, but since the -2 is just a constant, you can multiply that in and divide it.

So the integral would then become
S x (9-x^2)^(1/2) dx
= -1/2 S -2x (9-x^2)^(1/2) dx
and from here, you can do the reverse chain rule.

Hope that helps

 

[fullonoob]

You can do the reverse chain rule with any integral, as long as the derivative of the function is there.
But with this integral no.

If it was say, S x (9-x^2)^(1/2) dx instead, with the x at the front, then yes you could do the reverse chain rule, because the derivative of 9-x^2 is -2x, but since the -2 is just a constant, you can multiply that in and divide it.

So the integral would then become
S x (9-x^2)^(1/2) dx
= -1/2 S -2x (9-x^2)^(1/2) dx
and from here, you can do the reverse chain rule.

Hope that helps

lol thanks great help. I tried taking out - 1/2x LOL, wasn't sure what to do.

 

[cutemouse]

The 4U way would be to use the substitution x=sinθ or something.

 

[ilikebeeef]

The 4U way would be to use the substitution x=sinθ or something.

Use x=3sinθ.