3

The prompt is to find the volume of the solid which is described the equations and is bounded. $$x^2+y^2+z^2=9 $$ $$x^2-3x+y^2=0 $$ The first one is a sphere with radius 3, the shadow is on the y-x plane.

For the second on I tried using completing the squares. $$x^2-3x + y^2 =0 $$ $$x^2-3x+ 1/25 + y^2 = 1/25 $$ $$(x-1/5)^2 + y^2 = 1/25 $$

i dont know how to procede now. I also tried... $$x^2+y^2-3x = 0$$ $$r^2-3x = 0 $$ $$ r^2 = 3x$$ $$ r^2 = 3cos\theta$$ $$ r = \sqrt{3cos\theta} $$ $$ \int_0^3\int_0^{2\pi} \int_0^{\sqrt{3cos\theta}}x^2+y^2+z^9rdrd\theta dz$$ Please correct me if the method to get the radius, if its wrong? Im kinda new to calculus.

Prathik Gurudatt
  • 131

2 Answers2

2

the first is indeed a sphere of radius $3$ centered at the origin. But the second is a cylinder of radius $3/2$ centered at $(3/2,0,0)$ You have to fix the completion of squares:

$$x^2-3x+y^2=0\;;x^2-3x+9/4+y^2=9/4\;;(x-3/2)^2+y^2=9/4$$

But you try too to use cylindrical coordinates and it seems a better way. The equation for the cylinder is ok:

$r^2=3\cos\theta\;;r=\sqrt{3\cos\theta}$ (dropping the minus sign as it has to be $r\geq0$)

The equation for the sphere is $r^2+z^2=9$. Isolating $z$ to use for the integration limits, $z=\pm\sqrt{9-r^2}$

We are calculating the volume, so, we integrate only for the volume element $rdrd\theta dz$ (in fact, you tried to integrate the value of the square of the distance to the origing all over the volume; if we had to interpret that integral, it is the mean over the points of the region of their squared distance to the origin times the volume)

For the integration limits:

$-\pi/2\lt\theta\lt\pi/2\;;0\lt r\lt\sqrt{3\cos\theta}\;;-\sqrt{9-r^2}\lt z\lt\sqrt{9-r^2}$

$$V=\int_{-\pi/2}^{\pi/2}\int_0^\sqrt{3\cos\theta}\int_{-\sqrt{9-r^2}}^\sqrt{9-r^2}rdzdrd\theta$$

Rafa Budría
  • 7,552
  • does it have any function that is to be integrated? or is it just 1? – Prathik Gurudatt Jun 26 '17 at 13:50
  • Just $1$. Consider why $rdzdrd\theta$ is called often "volume element". If you inegrate the volume element over a region, you get the volume of this region! – Rafa Budría Jun 26 '17 at 13:52
  • Is that order a standard order you would recommend? And Can this be done with double integrals? – Prathik Gurudatt Jun 26 '17 at 14:12
  • There is no a standard order. For me the point is to find the easiest way to solve one variable as function or the other(s) to stabilish the intergration limits. In this case I think it's equivalent integrate first for $r$ (then $r=\pm\sqrt{9-z^2}$, etc) then for $z$ to the order I chose. Integrate first for $\theta$ needs the $\arcsin$ and I discarded it, but maybe it doesn't matter. Surely it can be done with double integrals, even a single one: e.g. taking slices parallel the the $xy$ plane and doing some trigonometry, but doing with double or triple depends on what you need to practice. – Rafa Budría Jun 26 '17 at 14:32
1

If you want integrate in cartesian coordinates, note that:

substituting $x^2+y^2=3x$ in the first equation we find that the limiting values for $z$ are given by $3x+z^2=9 $ that gives $$ -\sqrt{3(3-x)}<z<\sqrt{3(3-x)} $$

The circle in $x-y$ plane $x^2+y^2-3x=0$ gives: $y=\pm \sqrt{3x-x^2}$ so the limits for $y$ are: $$ -\sqrt{x(3-x)} <y<\sqrt{x(3-x)} $$ and, since $y$ must be real we have also the limits ofr $x$ $$ 0<x<3 $$ So, using the symmetry withe respect to the $x-y$ plane, the volume can be calculates as: $$ V=2\int_0^3\int_{-\sqrt{x(3-x)}}^{\sqrt{x(3-x)}}\int _0^{\sqrt{3(3-x)}}dzdydx $$

Emilio Novati
  • 64,377