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Find the volume above the x-y plane inside the cone $z=2-(x^{2}+y^{2})^{1/2}$ and inside the cylinder $(x-1)^{2} + y^{2}=1$

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Now using calculus this is actually a rather difficult integration using a Df matrix the bounds are rather un-intuitive to find, a simple substitution to polar actually doesn't make ur any life easier. ( you end up have to integrate $2-(1+r\cos(\theta)+r^{2})^{1/2}rdrd\theta$ in square cords u have $\int^{1} \int^{(1-(x-1)^{2})^{1/2}} (x^{2}+y^{2})^{1/2}dydx$ and although there is a formula in most calc text books to integrate this ingratiating again in x will be very ugly after the sub ( the upper bound isn't actually right but it has something like that in it each integral is evaluated 0 to upper bound)

$x-1=r\cos(\theta)$ and $y=r\sin(\theta)$ was the sub i used

curiously can anyone solve this without calculating a bound analytically merely by integrating up that surface in 1 step? ( no cheating and breaking it apart into 2 integrals ie this minus this or this plus this doubles are fine and i am fine with an integration by parts trick.) other then that any mathematical trickery is welcome.

Edit Understand the answer.

jimjim
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Faust
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2 Answers2

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Using Maple with $$\int _{0}^{2}\!\int _{-\sqrt {1- \left( x-1 \right) ^{2}}}^{\sqrt {1- \left( x-1 \right) ^{2}}}\! \left(2-\sqrt {{x}^{2}+{y}^{2}}\right) {dy}\,{dx} $$ it looks like the answer is $2 \pi - 32/9$, certainly not $2 \pi/3$.

EDIT: Actually, the easy way to do this is to use polar coordinates, noting that the equation of the cylinder is $r = 2 \cos(\theta)$ for $-\pi/2 < \theta < \pi/2$. The integral becomes $$ \int_{-\pi/2}^{\pi/2} \int_{0}^{2 \cos(\theta)} r (2 - r)\ dr \ d\theta$$

Robert Israel
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  • Can you show the choices for the cords you picked, please?

    When i sub into polar i get x= rcos thus x-1 = rcos +1 subbing into the other equation i get

    $2-(1+r\cos(\theta)+r^{2})^{1/2}rdrd\theta$

     – Faust Mar 13 '13 at 07:53
  • $x = r \cos(\theta)$, $y = r \sin(\theta)$, the equation of the cone is $z = 2 - \sqrt{x^2 + y^2} = 2 - r$. The cylinder is $(1-x)^2 + y^2 = 1$ which becomes $1 - 2 r \cos(\theta) + r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = 1$, and that simplifies to $r (- 2 \cos(\theta) + r) = 0$. – Robert Israel Mar 13 '13 at 07:58
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    I checked this in Mathematica separately. The cross section of the volume is the intersection of the circles $x^2+y^2<(2-z)^2$ and $(x-1)^2+y^2<1$. If we call this intersection $A(z)$ then the volume is

    $$\int_0^2 dz : A(z)$$ By using the Boole function in Mathematica, I verified that the answer is indeed $2 \pi - \frac{32}{9}$.

     – Ron Gordon Mar 13 '13 at 08:03
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Parametrization of the cylinder is $$ \begin{align*} x-1 &= \cos \theta \\ y &= \sin \theta \\ z &= r \end{align*} $$ To find the curve that's intersection of the cylinder and the cone, just substitute equations above to the equation of cone $$ z = 2-\sqrt{x^2+y^2} = 2-\sqrt{\left( \cos\theta+1\right)^2+\sin^2\theta} = 2-\sqrt{\cos^2\theta+\sin^2\theta+2\cos\theta+1} = \\ =2-\sqrt{2+2\cos\theta} = 2-\sqrt{4\cos^2\frac \theta 2} = 2\left(1\pm\cos\frac\theta 2\right) $$ So, polar coordinates of that curve is $$ \begin{align*} x &= \cos\theta+1\\ y &= \sin\theta \\ z &= 2\left(1\pm\cos\frac\theta 2\right) \end{align*} $$ It can be easily checked by Mathematica's ParametricPlot3D function

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So now, to find volume of this body, you need to find appropriate integration limits for the triple integral $$ V = \int_0^{z_0} \int_0^{\theta_0} \int_0^{r_0} rdr\,d\theta\,dz $$ where $z_0, \theta_0(z), r_0(z,\theta)$ are some functions, that need to be found based on the geometry of the body.

PS: I, personally, didn't find those limits yet, and this post is more of a hint, than a solution.

Kaster
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