Triple Integral: Convert from Cartesian to Cylindrical Coordinates

[daveyman]

Homework Statement

This is my last question about triple integrals in cylindrical coordinates.

Evaluate the integral by changing to cylindrical coordinates:

$$\int _{-3}^3\int _0^{\sqrt{9-x^2}}\int _0^{9-x^2-y^2}\sqrt{x^2+y^2}dzdydx$$

Homework Equations

In cylindrical coordinates, $$x^2+y^2=r^2$$ and $$x=r\cos{\theta}$$.

The Attempt at a Solution

My converted integral looks like this:

$$\int _0^{\pi }\int _0^{\sqrt{\frac{18}{1+\text{Cos}[\theta ]^2}}}\int _0^{9-r^2}r^2dzdrd\theta$$

This isn't quite right. Any ideas?

 

[tiny-tim]

Hi daveyman!

Hint: If |x| ≤ 3 and |y|² ≤ 9 - x², then … r what, and θ what?

 

[daveyman]

$$r=\sqrt{18-x^2}$$ which, in cylindrical coordinates is $$r=\sqrt{18-r^2\cos^2 \theta}$$.

So would the new integral be

$$\int _0^{\pi }\int _0^{\sqrt{18-r^2*\text{Cos}[\theta ]^2}}\int _0^{9-r^2}r^2dzdrd\theta$$?

 

[tiny-tim]

$$r=\sqrt{18-x^2}$$ which, in cylindrical coordinates is $$r=\sqrt{18-r^2\cos^2 \theta}$$.

So would the new integral be

$$\int _0^{\pi }\int _0^{\sqrt{18-r^2*\text{Cos}[\theta ]^2}}\int _0^{9-r^2}r^2dzdrd\theta$$?

No!

Get some graph paper and shade in all the points for which |x| ≤ 3 and |y|² ≤ 9 - x².​

 

[daveyman]

No!

Get some graph paper and shade in all the points for which |x| ≤ 3 and |y|² ≤ 9 - x².​

It is a circle at the origin with radius of three. So the integral will be

$$\int _0^{\pi }\int _0^3\int _0^{9-r^2}r^2dzdrd\theta$$

which yields the correct answer.

Thank you!

 

[tiny-tim]

It is a circle at the origin with radius of three. So the integral will be

$$\int _0^{\pi }\int _0^3\int _0^{9-r^2}r^2dzdrd\theta$$

which yields the correct answer.

Yup!

The moral of this … always draw the region first …

it's almost impossible to work out the limits without a diagram

(though of course, you must then prove them without a diagram … but that's much easier once the diagram has told you the answer! )​

 

[feldy90]

Hey- I know this is quite an old thread... But just wondering how you know that the dθ part is between 0 and pi??
Cheers