How to Integrate a Function with y^2 in It

[Mathpower]

Homework Statement

I would like to know how to integrate a function that has y^2 in it.

Homework Equations

eg: (x^2)/5 + (y^2) +900 = 0

The Attempt at a Solution

y = ((x^2)/5 - 900)^(1/2)
Area = ∫((x^2)/5 - 900)^(1/2)
Chain rule in reverse somehow?
= (((x^2)/5 - 900)^(3/2))/(1.5)
= No idea!

 

[oli4]

Hi Mathpower,
there must be something wrong in your relevant equations
the attempted solution is wrong respect to it, but just looking at your initial equation, you have a sum of squares + 900 = 0, that can't be solved since the left side is >=900

Cheers

 

[Mathpower]

Hi Mathpower,
there must be something wrong in your relevant equations
the attempted solution is wrong respect to it, but just looking at your initial equation, you have a sum of squares + 900 = 0, that can't be solved since the left side is >=900

Cheers

OK thank you for your help! What I am attempting to do is to learn how to do implicit differentiation in reverse.
Let me change the question:


1. Homework Statement
I would like to know how to integrate a function that has y^2 in it.


2. Homework Equations
eg: (x^2)/5 + (y^2) -1 = 0


3. The Attempt at a Solution
y = ((x^2)/5 - 1)^(1/2)
Area = ∫((x^2)/5 - 1)^(1/2)
Chain rule in reverse somehow?
= (((x^2)/5 - 1)^(3/2))/(1.5)
= No idea

 

[algebrat]

OK thank you for your help! What I am attempting to do is to learn how to do implicit differentiation in reverse.
Let me change the question:


1. Homework Statement
I would like to know how to integrate a function that has y^2 in it.


2. Homework Equations
eg: (x^2)/5 + (y^2) -1 = 0


3. The Attempt at a Solution
y = ((x^2)/5 - 1)^(1/2)
Area = ∫((x^2)/5 - 1)^(1/2)
Chain rule in reverse somehow?
= (((x^2)/5 - 1)^(3/2))/(1.5)
= No idea

it is hard to figure out what you wanted to do. did you want to calculate the area? where? in the first quadrant?

Try a change of variables [STRIKE]u^2=x^2/5[/STRIKE] (see comment below), then you'll have a nice circle. then you could do a change of variable to polar coordinates.

If you don't want to do that, then you can use trig subsitution when you have roots and squared terms.

 

[Infinitum]

(x^2)/5 + (y^2) -1 = 0

3. The Attempt at a Solution
y = ((x^2)/5 - 1)^(1/2)
Area = ∫((x^2)/5 - 1)^(1/2)

That should be,

$$y = \pm \sqrt{1-\frac{x^2}{5} }$$

So you would basically have two integrals to do,

$$\int y dy= \int \sqrt{1-\frac{x^2}{5} } dx$$

And,

$$\int y dy= \int -\sqrt{1-\frac{x^2}{5} } dx$$

Which one you have to integrate, and with what limits, depends on what you wish to find out, as algebrat observed.

 

[algebrat]

it is hard to figure out what you wanted to do. did you want to calculate the area? where? in the first quadrant?

Try a change of variables u^2=x^2/5, then you'll have a nice circle. then you could do a change of variable to polar coordinates.

If you don't want to do that, then you can use trig subsitution when you have roots and squared terms.

EGADS! I should have said u=x/√5 me thinks!

 

[Mathpower]

Oops I kind of mixed up what I wanted to learn with the actual test question that I had. My apologies!
If possible please show me how to do the indefinite integral (not the definite integral as my working had previously suggested) of (x^2)/5 + (y^2) -1 = 0

I am sure that I can pick up how to find the area after that. Oh...and let me know how I can post a message like Infinitum (ie: the math writing style...if you know what I mean)

 

[Infinitum]

Oops I kind of mixed up what I wanted to learn with the actual test question that I had. My apologies!
If possible please show me how to do the indefinite integral (not the definite integral as my working had previously suggested) of (x^2)/5 + (y^2) -1 = 0

I am sure that I can pick up how to find the area after that. Oh...and let me know how I can post a message like Infinitum (ie: the math writing style...if you know what I mean)

For area, integrate each of those expressions above, separately. Use the substitution, $u= \frac{x}{\sqrt{5}}$ You will have two answers, opposite in sign.

For only the indefinite integral, you can integrate the expression,

$$\int y^2 dy = \int (1-\frac{x^2}{5})dx$$

But this, you cannot use to find area.

As for the math, its called LaTeX. You can learn about it here : https://www.physicsforums.com/showthread.php?t=546968

 

[Mathpower]

WOW! Thank you so much for your help! Amazing that I am getting help from someone in USA!
oh...so:
Let: $$u^2 = (x^2 /5)$$
$$\sqrt{1-u^2}$$
Indefinite integral:
$$\int\sqrt{1x-u^2}$$
=$$2 (1x-u^2)^(-0.5) . (2u)$$
=$$4u/\sqrt(1x-u^2)$$
=$$(4x^2/5)/(1x-(x^2/5))^0.5$$
=4x^2/(5^(0.5)(5x-x^2))

I get the feeling I haven't done it correctly! The problem is I haven't been taught the whole integration topic yet...but I do know differentiation...and that doesn't really help in this situation!

 

[Infinitum]

Your steps after $\int \sqrt{1-u^2}$ aren't clear..

When you substitute a variable, you need to change the differential coefficient accordingly. You will have,

$$u = \frac{x}{\sqrt{5}}$$

$$du = \frac{dx}{\sqrt{5}}$$

Hence,

$$dx = \sqrt{5} du$$

Which you need to use in place of dx.

Also, once you reached $$\int \sqrt{1-u^2} du$$
You can do another substitution, $t=sin(u)$ and apply a similar technique as above. Then put back the t and u, so that you get x.

 

[HallsofIvy]

WOW! Thank you so much for your help! Amazing that I am getting help from someone in USA!

Why would you say that? I had not noticed that people from the USA refused to help others.

oh...so:
Let: $$u^2 = (x^2 /5)$$
$$\sqrt{1-u^2}$$
Indefinite integral:
$$\int\sqrt{1x-u^2}$$
=$$2 (1x-u^2)^(-0.5) . (2u)$$
=$$4u/\sqrt(1x-u^2)$$
=$$(4x^2/5)/(1x-(x^2/5))^0.5$$
=4x^2/(5^(0.5)(5x-x^2))

I get the feeling I haven't done it correctly! The problem is I haven't been taught the whole integration topic yet...but I do know differentiation...and that doesn't really help in this situation!

 

[algebrat]

Why would you say that? I had not noticed that people from the USA refused to help others.

Haha, maybe Mathpower just meant that they are far away from the USA, and it is amazing that the internet is bringing everyone together.

(You have assumed the negation is as you said.)

Perhaps we should start another thread on whether the USA is really helping anyone.

 

[Mathpower]

WOW! Thank you so much for your help! Amazing that I am getting help from someone in USA!
oh...so:
Let: $$u^2 = (x^2 /5)$$
$$\sqrt{1-u^2}$$
Indefinite integral:
$$\int\sqrt{1-u^2}$$
=2 (1x-u^2)^(-0.5) . (2u)
=4u/sqrt(1x-u^2)
=(4x^2/5)/(1x-(x^2/5))^0.5
=4x^2/(5^(0.5)(5x-x^2))

I get the feeling I haven't done it correctly! The problem is I haven't been taught the whole integration topic yet...but I do know differentiation...and that doesn't really help in this situation!

Oops I made a typo... I have corrected it in the above quote. Is it correct now?

 

[Mathpower]

Why would you say that? I had not noticed that people from the USA refused to help others.

Hold on...hold on... Talk about jumping to conclusions my friend!
I never even remotely meant neither suggested what you interpreted.
What I meant to say was that, it is quite amazing how someone 1000 's of kilometers away can help me!

 

[Infinitum]

Oops I made a typo... I have corrected it in the above quote. Is it correct now?

2 (1x-u^2)^(-0.5) . (2u)

Why the negative exponent? And how did you get the 'x' term from?? You substituted u to get rid of x... Also, you should read post #10 again

 

[Mathpower]

Ok ...woah ..what have I done!
I look back at my working now and I am shocked! I somehow managed to mix up differentiation with integration...ie: when differentiating we subtract 1 off the power, hence 0.5-1=-0.5
Lol!
But atleast I know why I did it and our class got taught integrating by substituting today! So I do know how to do it now...a very big thank you for your help.

Oh and just as a side note: I guess Americans really are so patriotic after all!

 

[dimension10]

Homework Statement

I would like to know how to integrate a function that has y^2 in it.

Homework Equations

eg: (x^2)/5 + (y^2) +900 = 0

The Attempt at a Solution

y = ((x^2)/5 - 900)^(1/2)
Area = ∫((x^2)/5 - 900)^(1/2)
Chain rule in reverse somehow?
= (((x^2)/5 - 900)^(3/2))/(1.5)
= No idea!

Let u=x/sqrt(5). Then use trig substitution.

 

[Infinitum]

Ok ...woah ..what have I done!
I look back at my working now and I am shocked! I somehow managed to mix up differentiation with integration...ie: when differentiating we subtract 1 off the power, hence 0.5-1=-0.5
Lol!
But atleast I know why I did it and our class got taught integrating by substituting today! So I do know how to do it now...a very big thank you for your help.

Good to know you got it clear now