How to solve a triple integral with a circle not centered at origin?

Use cylinder coordinates to calculate:$$\iiint\limits_{A}z\sqrt{x^2 + y^2}dA, \enspace A = \left\{(x,y,z):x^2 + (y - 1)^2 \leq 1,\: 0 \leq z \leq 2\right\}.$$

My works so far is this;

Switching to cylindrical coordinates we get:$$A = \left\{(r,\theta,z):0\leq r\leq 1,\: 0\leq \theta \leq 2\pi,\: 0\leq z \leq 2\right\}.$$ Now my book tells me that if you want the center in another point $(a,b,c)$, you should use the substitution: $$x = a + r\cos(\theta),\: y = b + r\sin(\theta),\: z = c + z.$$

With this in mind we change to cylindrical and add the bounderies (don't forget the jacobian).$$\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\int\limits_{0}^{2}zr\sqrt{r^2\cos^2(\theta) + (1 + r\sin(\theta))^2}dz\:dr\:d\theta \:=\:2\int\limits_{0}^{2\pi}\int\limits_{0}^{1} r\sqrt{r^2\cos^2(\theta) + (1 + r\sin(\theta))^2}dr \:d\theta.$$

Now this is the part where i get stuck, if i did my calculations right teh expression under the root becomes $r^2 + 1 + 2r\sin(\theta).$ I'm not sure where to go from here so any tips and tricks would be appreciated. I would very much like to see how this is done with this substitution, but alternative solution that leads me to the right answer would also be of great value. Finally, don't show me the calculations down to the answer as i would like to do that myself.

Your can use a simpler (?) transformation with cylindrical coordinates. Given that $$x^2+(y-1)^2=1$$ is a cylinder shifted 1 unit along the $$y$$ axis, you should not consider points in the $$y<0$$ plane ($$\pi \le \theta\le 2\pi$$), and the polar equation of the cylinder is not $$r=1$$ but $$r=2\sin\theta$$. Therefore, it should be $$A=\{(r,\theta,z)\;|\; 0 \le \theta \le \pi, 0 \le r \le 2\sin\theta, 0\le z \le 2\}$$

Once you have done that, the integral is easy and equals: $$\int_0^{\pi} \int_0^{2\sin\theta} \int_0^{2} z\, r\, r \; dz drd\theta = \frac{64}{9}$$

• I defined $A$ having the substitution (which shifts y) in mind which. That's why $A$ looks as a circle at the origin. Anyway your method is also great and something i've encountered before. Mar 4 '16 at 5:36
• @Kuifje I tried evaluating the last integral in the OP's post above. I was, regrettably, unsuccessful in attempting to find a simple closed form expression for the answer. However, I did get the value $7.11111$ when I typed it into my calculator. I double checked using Wolfram Alpha and got the same result. So it appears that the OP's definition of $A$ in cylindrical coordinates is, in fact, correct. Jul 8 '21 at 22:09
• thanks. I have updated my answer accordingly. Jul 9 '21 at 7:08

It's actually easier to evaluate the integral as it stands. I entered

$\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}z\sqrt{x^2+y^2} dxdydz$

in my calculator. It returned $64/9.$ You may want to look at this post where another person was having a very similar problem with a double integral.

As a general rule, a change of variables in a multiple integral is most effective when the same (or very similar) expressions occur in both the integrand and the region of integration.

• I disagree. A change in cylindrical coordinates makes the integral very easy here. Mar 3 '16 at 20:58
• @Kuifje Sorry, that was an exaggeration. The point is that this particular integral is really not that difficult in rectangular coordinates. Mar 3 '16 at 21:28
• I agree with you it is not difficult for a calculator, but it is (nearly) impossible by hand: how do you integrate $\int \sqrt{x^2+y^2} \;dx$ ? Mar 3 '16 at 21:33
• @Kuifje trigonometric substitution Mar 3 '16 at 21:35
• a trigonometric substitution is a change of variables Mar 3 '16 at 22:06