# triple integral over region

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

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.

I need to calculate :

$$\int\int\int x dxdydz$$

I think I can do this in two ways :

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

or

$$\int_{0}^{2}\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}}\int_{-x^2}^{2-x-y}xdzdydx$$

Questions:

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

• It seems all right. I don't think there is something simpler. In fact, the integrals in cartesians are not that hard. Feb 8, 2019 at 18:23
• 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 $$\int_0^{2 \pi} \int_0^1 (2 - x - y + x^2) x r \,dr d\theta = \int_0^{2 \pi} \int_0^1 (2 + 2 r^2 \cos^2 \theta) r \,dr d\theta.$$ Feb 8, 2019 at 19:47
• @Maxim isn't the radius $r$ from $0$ to $2\cos\theta$ ? How you do a step between your simplification?
– NPLS
Feb 8, 2019 at 19:51
• Different cylindrical coordinates (notice the offset in $x$). For the simplification, expand the integrand and find the terms containing odd powers of $\sin$ or $\cos$. Feb 8, 2019 at 19:58
• @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 $x^2+y^2=2x$ --> $r^2=2r\cos\theta$ --> $r=\cos\theta$ ?
– NPLS
Feb 8, 2019 at 20:14

$$I=\int_0^2 dx \int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}}dy \int_{-x^2}^{2-x-y} dz x$$ $$= \int_0^2 dx \int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}}dy (2-x-y+x^2) x$$ $$= \int_0^2 dx x(2y-xy-\frac12y^2+x^2y) \mit_{y=-\sqrt{2x-x^2}}^{y=\sqrt{2x-x^2}}$$ $$= \int_0^2 dx [4x-2x^2+2x^3]\sqrt{2x-x^2}$$ 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.