# Integral of a paraboloid using polar coordinates

#### Hckyplayer8

##### Full Member

Hello. I was working on this problem and got stuck after converting the integral to polar coordinates and integrating theta. Because the function converts to r^2

the integral becomes

My question involves the next step which tells me to assume $1+4r^2=t$
My question is, why? Is this just an arbitrary variable that just so happens to be t? Or does it mean more?

View attachment 37566

Hello. I was working on this problem and got stuck after converting the integral to polar coordinates and integrating theta. Because the function converts to r^2

the integral becomes View attachment 37567

My question involves the next step which tells me to assume $1+4r^2=t$
My question is, why? Is this just an arbitrary variable that just so happens to be t? Or does it mean more?
Are you saying that you are following step-by-step instructions, without understanding the steps?

I haven't checked the work so far, but at this step, you are being told to make a substitution. People often use the variable u and call it a "u-substitution", but you can use t if you wish.

Do you know about integration by substitution, and how to recognize when it will work? (Hint: if t = 4r^2 + 1, then dt = 8r, which almost appears in the integral.)

It is not an "assumption", and is not arbitrary.

View attachment 37566

Hello. I was working on this problem and got stuck after converting the integral to polar coordinates and integrating theta. Because the function converts to r^2

the integral becomes View attachment 37567

My question involves the next step which tells me to assume $1+4r^2=t$
My question is, why? Is this just an arbitrary variable that just so happens to be t? Or does it mean more?
How did you get this integral? Can you post your intermediate steps ?
Thank you.

How did you get this integral? Can you post your intermediate steps ?
Thank you.
Sure can. Thanks!

Are you saying that you are following step-by-step instructions, without understanding the steps?

I haven't checked the work so far, but at this step, you are being told to make a substitution. People often use the variable u and call it a "u-substitution", but you can use t if you wish.

Do you know about integration by substitution, and how to recognize when it will work? (Hint: if t = 4r^2 + 1, then dt = 8r, which almost appears in the integral.)

It is not an "assumption", and is not arbitrary.

That's what I wanted to make sure of. That it was just a simple u sub and didn't stand for something else.

If it is just a u-sub, shouldn't the substitution be squared since one of the variables that the substitution stands for is squared?

That's what I wanted to make sure of. That it was just a simple u sub and didn't stand for something else.

If it is just a u-sub, shouldn't the substitution be squared since one of the variables that the substitution stands for is squared?
Just try it and see! Replace 4r^2 + 1 with t, or u, or whatever, and r dr with 1/8 dt or 1/8 du, and see how it works. If you think you should do something else, try that and see if it works. Then show us, and we can discuss it.

My question is, why? Is this just an arbitrary variable that just so happens to be t? Or does it mean more?
Just one of those tricks one learns after many exercises. When I see [imath]r dr[/imath] my first thought is [imath]\frac{1}{2} d (r^2)[/imath], or [imath]t=r^2[/imath]. Then I see [imath]r^2[/imath] under the radical and remember that 'd' of constant is 0, so why not use [imath]t=4r^2+1[/imath] to reduce the radical to a simple power of [imath]u[/imath].

Bravo OP. You have done the hard part awesomely and got stuck in the easiest part. I am not surprised that you got stuck in that part. Usually, you have done hundreds of u-substitutions, but you have done them in the cartesian coordinate. And now when the problem gave you the hint to work in the polar coordinate you probably forgot or got confused that the same rules can be applied. Don't look at what coordinate you're working in, just apply the rules and tricks of integration once you get the opportunity to use them.

You were just so lucky that the integral was not like this:

[imath]\displaystyle 2\pi\int_{0}^{a} \sqrt{4r^2 + 1} \ dr[/imath]