# Thread: Method for converting integral:cylindrical coord to rectangular coord

1. ## Method for converting integral:cylindrical coord to rectangular coord

I want to convert integrals $\int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta$ from cylindrical coordinate
to rectangular coordinate( $x,y$ and $z$). Is there an easy way to convert these integrals from cylindrical
coordinate to rectangular coordinate?

What is the best method for this conversion?

The answer for the above problem is: $\int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx + \int_1^{\sqrt{3}} \int_0^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx$

Thanks.

2. For converting
$\int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta$
integral from cylindrical coordinate to rectangular coordinate first i find the limits(upper/lower boundaries of x,y and z).

We know that (from the integral)
$
z = \sqrt{4-r^2}
$

so we can say the upper limit of $z$ is:
$
z = \sqrt{4-x^2-y^2}
$

and lower limit of $z$ is:
$
z = 1
$

Also we know from the integral that
$
r= \sqrt{3}
$

So we can say:
$
\sqrt{(r \sin(\theta))^2 +(r \cos(\theta))^2} = \sqrt{3}
$

And:
$
y^2 + x^2 = 3
$

As a result the upper limit of $y$ is:
$
y = \sqrt{3 - x^2}
$

Again from the integral $r = 1$
Using the same procedure we can say lower limit of $y$ is: $y = \sqrt{1 - x^2}$

Upper limit of $x$ is
$
x = \sqrt{3}
$

and lower limit of $x$ is
$
x = 1
$

Now my result is:
$
\int_1^{\sqrt{3}} \int_{\sqrt{1 - x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4-x^2-y^2}} \,\, z^2yx\,\, dz\,\, dy\,\, dx
$

But the answer is different. It is:
$\int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx + \int_1^{\sqrt{3}} \int_0^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx
$

Why the answer is different? Can anyone kindly point out where i did wrong?

3. If I convert $\int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta$

to rectangular coordinate can it be equal to

$\int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx + \int_1^{\sqrt{3}} \int_0^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx
$
?

Please a simple yes or no will do.

4. $z= \sqrt{4- r^2}$ is almost the same as $z^2= 4- r^2= 1- x^2- y^2$ or $x^2+ y^2+ z^2= 4$, the sphere with center at (0, 0, 0) and radius 2. "Almost" because, of course, the square root is positive so this is actually the hemi-sphere above the xy-plane. The lower limit on z is just z= 1.

Taking $\theta$ from 0 to $\pi/2$ and r from 1 to $\sqrt{3}$ means that we, in the x, y plane, we are integrating between the circles $x^2+ y^2= 1$ and $x^2+ y^2= 3$ in the first quadrant.

If we want to do this integral in the order $\int\int\int dzdydx$ we will need to break it into two parts. For x between 0 and 1, the lower boundary is the smaller circle, $x^2+ y^2= 1$ or $y= \sqrt{1- x^2}$, since we are in the first quadrant, and the upper boundary is the larger circle, $x^2+ y^2= 3$ or $y= \sqrt{3- x^2}$ since we are in the first quadrant. For x between 1 and $\sqrt{3}$, the upper boundary is still $y= \sqrt{3- x^2}$ but the lower boundary is y= 0. In either integral, z goes from 1 to $\sqrt{4- x^2- y^2}$.

The limits of integration will be $\int_{x=0}^1\int_{y= \sqrt{1- x^2}}^{\sqrt{3- x^2}}\int_{z=1}^{\sqrt{4- x^2- y^2}} dz dy dx+ \int_{x= 1}^{\sqrt{3}}\int_{y= 0}^{\sqrt{3- x^2}}\int_{z= 1}^{\sqrt{4- x^2- y^2}} dz dy dx$.

Your integrand is $r^3 sin(\theta)cos(\theta)z^2 dz dr d\theta$ but the differential of volume in cylindrical coordinates is $r dz dr d\theta$ so we can write that as $r^2 sin(\theta)cos(\theta)z^2 (r dz dr d\theta)= (r sin(\theta))(r cos(\theta)) z^2 dz dy zr= xyz^2 dz dy dx$

Putting that all together,
$\int_{x=0}^1\int_{y= \sqrt{1- x^2}}^{\sqrt{3- x^2}}\int_{z=1}^{\sqrt{4- x^2- y^2}} xyz^2 dz dy dx+ \int_{x= 1}^{\sqrt{3}}\int_{y= 0}^{\sqrt{3- x^2}}\int_{z= 1}^{\sqrt{4- x^2- y^2}} xyz^2 dz dy dx$

which is just what you have! Well done!

5. HallsofIvy, all i needed is your simple explanation. And thank you for that.