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NIn the following example I have been asked to find the volume V of the solid bounded by the sphere $ x^2 + y^2 +z^2 = 2 $ and the paraboloid $ x^2 + y^2 = z $ by using triple integration.

I am not quite sure how to set up this triple integration and what each of the integrands should be.

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Note $z=x^2+y^2>0$, therefor $$V=\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\int_{x^2+y^2}^{\sqrt{2-x^2-y^2}}dzdydx$$ Now apply cylinder coordinate.

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    $\begingroup$ I am not quite sure what cylinder coordinates are yet $\endgroup$ – mp12345 Jul 24 '16 at 23:10
  • $\begingroup$ Surely You should use cylindrical coordinates. $\endgroup$ – Behrouz Maleki Jul 25 '16 at 0:12
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    $\begingroup$ The OP said that they don't know what these are. No need to berate. @javahelp: While Cartesian coordinates describe an $x$, $y$, and $z$ offsets from a fixed origin, cylindrical coordinates describe a height $z$ above a fixed plane, and a radius $r$ and angle $\theta$ within that plane. Imagine drawing a cylinder centered at the origin and passing through your point; you can describe the point's position by its height, radius, and angle. Here's an article about integrating in cylindrical coordinates./ $\endgroup$ – wchargin Jul 25 '16 at 5:07
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It is comfortable to describe your solid in cylindrical coordinates since both the sphere and the paraboloid have the $z$-axis as an axis of symmetry (both surfaces are surfaces of revolution around the $z$-axis).

In cylindrical coordinates, the sphere is described by $\rho^2 + z^2 = 2$ and the paraboloid is described by $\rho^2 = z$. The solid bounded between them is described by $\rho^2 \leq z \leq \sqrt{2 - \rho^2}$ and so

$$ V = \int_{0}^1 \int_{0}^{2\pi} \int_{\rho^2}^{\sqrt{2 - \rho^2}} \rho \, dz \, d\theta \, d\rho = 2\pi \int_0^{1} \rho \left( \sqrt{2 - \rho^2} - \rho^2 \right) \, d\rho = 2\pi \left( \frac{1}{2} \int_1^2 \sqrt{u} \, du - \int_0^{1} \rho^3 \right) = 2\pi \left( \frac{2^{\frac{3}{2}} - 1}{3} - \frac{1}{4}\right).$$

where we used the substitution $u = 2 - \rho^2, \, du = -2\rho$ for the first integral.

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    $\begingroup$ okay so 4pi/3 2^3/2 is the final answer? $\endgroup$ – mp12345 Jul 24 '16 at 23:26
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    $\begingroup$ okay so this answer is wrong then? $\endgroup$ – mp12345 Jul 24 '16 at 23:47
  • $\begingroup$ @AhmedHussein Edited, thanks for the correction! $\endgroup$ – levap Jul 25 '16 at 0:27
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    $\begingroup$ Okay so you have corrected your mistake in that edit? $\endgroup$ – jh123 Jul 25 '16 at 0:57

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