Integral Gives Different Answers on Polar Coordinates

[Flemish]

 

Hello, 

I have been stuck on this Math problem and wanted some help. This is the formula for finding out the surface area:

Using this formula, you should arrive on something like this (the outer limits are pi/4 and 0, the program didn't let me input).

 

 

Focusing on the inner integral, by rearranging you arrive on this

If you integrate this, the answer will always be zero:

 

The answer to this always has to be zero, because inputting the limits inside the square root would give zero. However, if you switched to polar coordinates, you'd arrive on a different answer:

 

 

Which would give you an answer of 2(pi)(a^2). I searched around and people were saying that answers mean different things in different coordinate systems. I understand this, but if region R is a constant area throughout the different coordinate systems, then an answer of zero in the cartesian plane would suggest that the surface area is zero. Meaning that R hasn't have any surface area (assuming R is a constant area through coordinate systems), then no matter the coordinate system, the answer should be the same. Where did I go wrong?

[Genady]

5 minutes ago, Flemish said:

the outer limits are pi/4 and 0

The first stop: I think these limits are wrong.

[Lorentz Jr]

1 hour ago, Flemish said:

The answer to this always has to be zero

The y in the denominator changes sign. You're integrating over y, not x.

Your inner integral is wrong. Try again.

Start with [math]b^2 \equiv a^2 - x^2[/math].

Edited January 1, 2023 by Lorentz Jr

[Flemish]

7 hours ago, Genady said:

The first stop: I think these limits are wrong.

Sorry, I meant from a to -a

7 hours ago, Lorentz Jr said:

The y in the denominator changes sign. You're integrating over y, not x.

Your inner integral is wrong. Try again.

Start with b2≡a2−x2 .

Sir, I don't get what you mean by start with b^2 = a^2- x^2, are you referring to the limits or a substitution?  If its a substitution then how would it fix our problem?

If your talking about the limits, here is how I got them:

My logic is that because its a circle, the y-values have to be given by this.

[Lorentz Jr]

26 minutes ago, Flemish said:

If its a substitution then how would it fix our problem?

-b < y < b

[Flemish]

Just now, Lorentz Jr said:

-b < y < b

That wouldn't fix the problem because the inside the square root will still be zero

[Lorentz Jr]

2 minutes ago, Flemish said:

the inside the square root will still be zero

That part is wrong. You need to go back and redo the inside integral in y.

Edited January 1, 2023 by Lorentz Jr

[Flemish]

6 minutes ago, Lorentz Jr said:

That part is wrong. You need to go back and redo the inside integral.

what is wrong with it, I double checked

[Lorentz Jr]

You're integrating f^-1/2 dy, not f^-1/2 df.

 

[studiot]

11 minutes ago, Lorentz Jr said:

You're integrating f^-1/2 dy, not f^-1/2 df.

 

 

8 hours ago, Genady said:

The first stop: I think these limits are wrong.

Very Helpful, guys.  +1

[Lorentz Jr]

No more OP....

df = -2y dy, so f^-1/2 df would be -2 f^-1/2 y dy.

But the integral in the problem is f^-1/2 dy, which would be - f^-1/2 df / 2y.

So still not f^-1/2 df.

Edited January 2, 2023 by Lorentz Jr

[Flemish]

On 1/2/2023 at 8:45 AM, Lorentz Jr said:

No more OP....

df = -2y dy, so f^-1/2 df would be -2 f^-1/2 y dy.

But the integral in the problem is f^-1/2 dy, which would be - f^-1/2 df / 2y.

So still not f^-1/2 df.

Thank you sir! I understand where your coming from. Sorry my brain wasn't working before.