# Change the order of integration in Spherical coordinate and Cylindrical Coordiate

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

• Have you sketched? Have you figured out the bounds for any of the orders of integration in cylindrical / spherical coordinates? Please edit the question showing what you have already done / attempted. Commented Nov 14, 2020 at 17:55
• @MathLover I just edit the post and the picture has what I got so far, I have no idea how to do the question b and c
– Boba
Commented Nov 14, 2020 at 18:25
• OK , there are some mistakes in your integral setup. Pls see my answer. Next time on, pls try and share your workings in the question body itself instead of an image. People may just downvote if they do not see your effort in the body of the question. Commented Nov 14, 2020 at 20:30

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 $$\rho$$.