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Find the volume of the solid enclosed by the paraboloids $z = 1-x^2-y^2$ and $z = -1 + (x-1)^2 + y^2$.

Using triple integrals, it is known that $V = \iiint_R \mathrm dx\,\mathrm dy\,\mathrm dz$, and I will have to change variables. But I can't just say that $r^2 = x^2 + y^2$, because the second paraboloid has a "$(x-1)^2$" term. How do I change variables without making a mess?

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    $\begingroup$ A very similar question is mentioned here. Take a look at that. :) $\endgroup$ Commented Dec 22, 2015 at 13:21

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Let $$x_1=x-\dfrac12;\quad z_1=z-x_1,$$ then $$z_1^{(1)} = \dfrac34-x_1^2-y^2,\quad z_1^{(2)} = -\left(\dfrac34-x_1^2-y^2\right).$$ We can see:

  1. $z_1^{(1)} = z_1^{(2)}=0$ when $x_1^2+y^2=\dfrac34$
  2. $z_1^{(1)} > z_1^{(2)}$ when $x_1^2+y^2<\dfrac34$ - finite volume
  3. $z_1^{(1)} < z_1^{(2)}$ when $x_1^2+y^2>\dfrac34$ - infinite volume.
    Choosing the finite volume, we have: $$y\in\left(-\dfrac{\sqrt3}2,\dfrac{\sqrt3}2\right),$$ $$x_1\in\left(-\sqrt{\:\dfrac34-y^2},\:\sqrt{\:\dfrac34-y^2}\right),$$ $$z_1\in\left(0,\:\dfrac34-x_1^2-y^2\right)$$ Reshuffle limited shift coordinates and in accordance with the principle of Cavalieri areas and volumes do not change, so that $$V=\int\limits_{-\dfrac{\sqrt3}2}^{\dfrac{\sqrt3}2}dy\int\limits_{-\sqrt{\dfrac34 - y^2}}^{\sqrt{\dfrac34 - y^2}}dx_1\int\limits_0^{\dfrac34-x_1^2-y^2}dz_1 = \int\limits_{-\dfrac{\sqrt3}2}^{\dfrac{\sqrt3}2}dy\int\limits_{-\sqrt{\dfrac34 - y^2}}^{\sqrt{\dfrac34 - y^2}}\left(\dfrac34-x_1^2-y^2\right)dx_1 = {4\int\limits_0^{\dfrac{\sqrt3}2}dy\int\limits_0^{\sqrt{\dfrac34 - y^2}}\left(\dfrac34-y^2-x_1^2\right)dx_1} = 4\int\limits_0^{\dfrac{\sqrt3}2}\left.x_1\left(\dfrac34-y^2-\dfrac13x_1^2\right)\right|_{x_1=0}^{\sqrt{\dfrac34 - y^2}}dy = {\dfrac83\int\limits_0^{\dfrac{\sqrt3}2}\left(\dfrac34-y^2\right)^{\dfrac32}dy}$$ Using substitution $y=\dfrac{\sqrt3}2\sin t,\: dy = \dfrac{\sqrt3}2\cos t dt:$ $$V = \dfrac32\int\limits_0^{\dfrac{\pi}2}\cos^4tdt = \dfrac38\int\limits_0^{\dfrac{\pi}2}(1+\cos2t)^2dt = \dfrac38\int\limits_0^{\dfrac{\pi}2}\left(1+2\cos2t+\dfrac12(1+\cos4t)\right)dt = \dfrac3{16}\int\limits_0^{\dfrac{\pi}2}(3 + 4\cos2t + \cos4t)dt = \left.\dfrac3{16}(3t + 2\sin2t + \dfrac14\sin4t)\right|_0^{\dfrac{\pi}2} = \dfrac9{32}\pi.$$
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Since $z=1-x^2-y^2$ is a downward paraboloids and $z=-1+(x-1)^2+y^2$ is upward , the limits for $z$ are:

$$ -1+(x-1)^2+y^2 \le z \le 1-x^2-y^2 $$

The projection on the $xy$ plane of the curve of intersection between the two paraboloids is the circumference:

$$ x^2+y^2-x-\frac{1}{2}=0 $$

so the limits for $y$ are: $$-\sqrt{\frac{1}{2}+x-x^2} \le y \le \sqrt{\frac{1}{2}+x-x^2} $$

that has real solution if $x$ in the limits $$ \frac{1-\sqrt{3}}{2}\le x \le \frac{1+\sqrt{3}}{2} $$

so the volume is given by:

$$ \int_{\frac{1-\sqrt{3}}{2}}^{\frac{1+\sqrt{3}}{2}} \int_{-\sqrt{\frac{1}{2}+x-x^2}}^{\sqrt{\frac{1}{2}+x-x^2}} \int_{[-1+(x-1)^2+y^2]}^{[1-x^2-y^2]} dzdydx $$

Since the problem has not an axis of symmetry, it is unlikely that the use of cylindrical coordinates gives a simpler integration.

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