$D=\{x^2-2x+y^2\le 0 ,-x^2\le z \le 2-x-y\}$

My attempt:

The first one is a shifted cylinder with a radius : $x^2+y^2=2x$ which in polar coordinate should be $r=2cos\theta$.

Cylinder parametrization :

$$\begin{cases} x=1+r\cos\theta \\ y=r\sin\theta \\ z=z \end{cases} $$

I chose $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ becouse the "shadowed" region is a circle tangent to the z-y plane.

enter image description here

I need to calculate :

$\int\int\int x dxdydz$

I think I can do this in two ways :

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2cos\theta}\int_{-r^2cos^2\theta}^{2-rcos\theta-rsin\theta}r^2cos\theta dzdrd\theta$




The problem is that they seem to be too hard, maybe something will simplify... but I really can't figure out how.

So my question is : Are those integral right? And if So can you give me a hint on how to solve it?

Book Answer : $\frac{3\pi}{2}$

  • $\begingroup$ It seems all right. I don't think there is something simpler. In fact, the integrals in cartesians are not that hard. $\endgroup$ Feb 8, 2019 at 18:23
  • $\begingroup$ Both of your triple integrals are correct. You can simplify things by taking the cylindrical coordinates $(x, y, z) = (1 + r \cos \theta, r \sin \theta, z)$ and using the fact that $\int_0^{2 \pi} \sin^m \theta \cos^n \theta \,d\theta$ is non-zero only if both $m$ and $n$ are even. You'll get $$\int_0^{2 \pi} \int_0^1 (2 - x - y + x^2) x r \,dr d\theta = \int_0^{2 \pi} \int_0^1 (2 + 2 r^2 \cos^2 \theta) r \,dr d\theta.$$ $\endgroup$
    – Maxim
    Feb 8, 2019 at 19:47
  • $\begingroup$ @Maxim isn't the radius $r$ from $0$ to $2\cos\theta$ ? How you do a step between your simplification? $\endgroup$
    – NPLS
    Feb 8, 2019 at 19:51
  • $\begingroup$ Different cylindrical coordinates (notice the offset in $x$). For the simplification, expand the integrand and find the terms containing odd powers of $\sin$ or $\cos$. $\endgroup$
    – Maxim
    Feb 8, 2019 at 19:58
  • $\begingroup$ @Maxim I don't understand why $\theta$ is between $0$ and $2\pi$ even though the circle has an offset. what's wrong with writing $x^2+y^2=2x$ --> $r^2=2r\cos\theta$ --> $r=\cos\theta$ ? $\endgroup$
    – NPLS
    Feb 8, 2019 at 20:14

1 Answer 1


$$ I=\int_0^2 dx \int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}}dy \int_{-x^2}^{2-x-y} dz x $$ $$ = \int_0^2 dx \int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}}dy (2-x-y+x^2) x $$ $$ = \int_0^2 dx x(2y-xy-\frac12y^2+x^2y) \mit_{y=-\sqrt{2x-x^2}}^{y=\sqrt{2x-x^2}} $$ $$ = \int_0^2 dx [4x-2x^2+2x^3]\sqrt{2x-x^2} $$ Here $$ \int x\sqrt{2x-x^2} = -\frac13 (2x-x^2)^{3/2}-\frac12 (1-x)(2x-x^2)^{1/2}+\frac12 \arcsin(x-1); $$ $$ \int x^2\sqrt{2x-x^2} = -\frac{x}{4} (2x-x^2)^{3/2}-\frac{5}{12}(2x-x^2)^{3/2}-\frac58 (1-x)(2x-x^2)^{1/2}+\frac58\arcsin(x-1); $$ $$ \int x^3\sqrt{2x-x^2} = -[x^2/5-7x/20-7/12] (2x-x^2)^{3/2}-\frac78 (1-x)(2x-x^2)^{1/2}+\frac78\arcsin(x-1); $$ Combined $$ \int (4x-2x^2+2x^3)\sqrt{2x-x^2} dx = [-2x^2/5 -x/5-5/3](2x-x^3)^{3/2}-\frac52(1-x)(2x-x^2)^{1/2}+\frac52\arcsin(x-1), $$ so $$ I=\int_0^2 (4x-2x^2+2x^3)\sqrt{2x-x^2} dx = \frac52[\arcsin 1-\arcsin(-1)]=5\arcsin 1 = \frac52 \pi. $$ This is NOT the book's answer. I suspect the book's answer is wrong.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.