[Math] Change the order of integration in Spherical coordinate and Cylindrical Coordiate

calculusmultivariable-calculus

Let D be the region bounded below by the plane z = 0, above by the sphere $x^2 +y^2 +z^2 =4$, and on the sides by the cylinder $x^2 + y^2 = 1$. Set up the triple integrals in cylindrical coordinates that enter preformatted text heregive the volume of D using the following orders of integration.

a. $dzdrdu$ b. $drdzdu$ c. $dudzdr$

Let D be the region in Exercise 33. Set up the triple integrals in spherical coordinates that give the volume of D using the follow- ing orders of integration.

a. $d\rho d\phi d\theta $ b. $d\phi d\rho d\theta$

And I'm also wondering if it is correct that we can switch the order of integration directly if the bounds are all constant and independent on each other

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Best Answer

There is a mistake in your cylindrical coordinates working and your spherical coordinate is wrong. Bounds of $r$ is from $0$ to $1$ and not from $0$ to $2$. None of the discs that you take will have radius more than $1$. Please also note the upper bound of $z = \sqrt{4-r^2} = 2 \,$ when $r = 0$ and $z = \sqrt{3} \,$ when $r = 1$.

So, $V = \displaystyle \int_{0}^{2\pi} \int_{0}^1 \int_{0}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$

Now in spherical coordinates, for simplicity, your integral will be two parts - one which is spherical cone between $0 \leq \phi \leq \frac{\pi}{6}$ and then rest of the part of the cylinder which is between $\frac{\pi}{6} \leq \phi \leq \frac{\pi}{2}$.

$V = \displaystyle \int_{0}^{2\pi} \int_{0}^{\pi/6} \int_{0}^{2} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \, \, + \int_{0}^{2\pi} \int_{\pi/6}^{\pi/2} \int_{0}^{csc (\phi)} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

You should check your sketch to understand why the second integral goes from $0$ to $\csc \phi$ (if you draw a line from the center (origin in this case) to any point on the cylinder, it is hypotenuse of a right angled triangle with angle $\phi$ to the $z$ axis and distance along $xy$ axis being $1$. So $\rho$ is $1/\sin \phi$.)

Now for the change of order of integral in cylindrical coordinates, please note $(c)$ is pretty straightforward. Now for (b), you need to use your sketch to think how to go about it - you may have to split it into two parts ($0 \leq r \leq 1, 1 \leq r \leq \sqrt{4-z^2}$). Think similarly for the other integral in spherical coordinates - how to represent $\phi$ in in terms of $\rho$.

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