# Yet another integration

#### paw

Have tried to solve this one with substitution, but didn't quite get there (according to Wolfram Alpha):

$$\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \left( \sqrt{R^2-x^2-y^2} \right) \, dy$$

Care to help?

-paw

#### topsquark

Math Team
Have tried to solve this one with substitution, but didn't quite get there (according to Wolfram Alpha):

$$\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \left( \sqrt{R^2-x^2-y^2} \right) \, dy$$

Care to help?

-paw
As far as the integration is concerned R and x are constants. So let $$\displaystyle a^2 = R^2 - x^2$$. Then your integral becomes
$$\displaystyle \int_{-a}^a \sqrt{a^2 - y^2}~dy$$

How do you attack this kind of integral in general?

-Dan

• 2 people

#### paw

As far as the integration is concerned R and x are constants. So let $$\displaystyle a^2 = R^2 - x^2$$. Then your integral becomes
$$\displaystyle \int_{-a}^a \sqrt{a^2 - y^2}~dy$$

How do you attack this kind of integral in general?

-Dan

Integration of square root of (a^2- x^2)dx - YouTube

Even more confused.

#### MarkFL

Integration of square root of (a^2- x^2)dx - YouTube

Even more confused.
Rather than post a link to a video, why not take what you gleaned from the video and post your attempt so that those helping do not have to follow a link, load a video, watch the video, and then guess what you got from it and tried? • 1 person

#### paw

The integral is the exact same integral topsquark presented to me, and should therefore give the same solution. The derivation was quite extensive and more tricky than I expected it to be.

The result he ends up with is: $\frac{a^2}{2} \arcsin(\frac{y}{a})+\frac{y}{2} \sqrt{a^2-y^2}$

In my case:
$\biggr[\frac{a^2}{2} \arcsin(\frac{y}{a})+\frac{y}{2} \sqrt{a^2-y^2} \biggr]_{-a}^{a} = a^2\arcsin(\frac{a}{a})+a \sqrt{a^2-a^2} = a^2\arcsin(1)$

This does not look like the answer I end up with when using Wolfram Alpha.

Last edited by a moderator:

#### topsquark

Math Team
The integral is the exact same integral topsquark presented to me, and should therefore give the same solution. The derivation was quite extensive and more tricky than I expected it to be.

The result he ends up with is: $\frac{a^2}{2} \arcsin(\frac{y}{a})+\frac{y}{2} \sqrt{a^2-y^2}$

In my case:
$\biggr[\frac{a^2}{2} \arcsin(\frac{y}{a})+\frac{y}{2} \sqrt{a^2-y^2} \biggr]_{-a}^{a} = a^2\arcsin(\frac{a}{a})+a \sqrt{a^2-a^2} = a^2\arcsin(1)$

This does not look like the answer I end up with when using Wolfram Alpha.
As $$\displaystyle \arcsin(1) = \frac{\pi}{2}$$ I'd say you've got it correct. -Dan

Last edited by a moderator:
• 2 people

#### paw

As $$\displaystyle \arcsin(1) = \frac{\pi}{2}$$ I'd say you've got it correct. -Dan
That was great! Just finished the rest of the integral, and got it right Thank you topsquark.

#### topsquark

Math Team
That was great! Just finished the rest of the integral, and got it right Thank you topsquark.
I live to serve. :spin:

-Dan

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