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A body is surrounded by its lateral faces:

$$z(x,y) = h \left(1 - \left(\frac{x}{a}\right)^2 - \left(\frac{y}{b}\right)^2 \right)$$

and

$$z(x,y)=0$$

It should be a paraboloid, right? How can I calculate its volume via integration over $x$ and $y$ in Cartesian coordinates?

Thanks a lot in advance!

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We see that $z$ should be between $0$ and $\hat{z}(x, y)=h\left(1-\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2\right)$, so the integral in $z$ will be $$\int_0^{\hat{z}(x,y)}\mathrm{d}z$$ We can see zhat $\hat{z}(-a,0)=\hat{z}(a, 0)=0$, so the integral in $x$ will be $$\int_{-a}^{a}\mathrm{d}x$$ And finally, if $x$ is fixed, then we want to find $2$ numbers, for which $\hat{z}(x,y_1)=\hat{z}(x,y_2)=0$, i.e. we want to solve $$0=1-\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2$$ $$1=\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2$$ $$\hat{y}_{\pm}(x)=\pm b\sqrt{1-\left(\frac{x}{a}\right)^2}$$ And we will get that the integral in $y$ will be $$\int_{\hat{y}_{-}(x)}^{\hat{y}_{+}(x)} \mathrm{d}y$$ Combinating them, the final integral is $$V=\int \mathrm{d}V=\int_{-a}^{a}\mathrm{d}x\left(\int_{\hat{y}_{-}(x)}^{\hat{y}_{+}(x)} \mathrm{d}y\left(\int_0^{\hat{z}(x,y)}\mathrm{d}z\right)\right)$$

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In the case $a,\,b,\,h>0$ we seek $$V:=\int_{0}^{h(1-(x/a)^2-(y/b)^2)}dz\int_{(x/a)^2+(y/b)^2\le 1}dxdy,$$so the substitution $x=au,\,y=bv,\,z=hw$ gives$$\frac{V}{abh}=\int_0^{1-u^2=v^2}dw\int_{u^2+v^2\le 1}dudv=\int_{u^2+v^2\le 1}(1-u^2-v^2)dudv.$$Finally, write $u=r\cos s,\,v=r\sin s$ so $$\frac{V}{abh}=\int_0^{2\pi}ds\int_0^1dr\: r(1-r^2)=\frac{\pi}{2}.$$Of course, if any of $a,\,b$ is negative the same volume must result, while if $h<0$ we get a negative integral but would still say a positive volume was bound. The desired volume is therefore $\frac{\pi}{2}|abh|$.

As a sanity check, note that square-rooting the upper bound on $|z/h|$ increases the volume to obtain a hemiellipsoid, with the $\pi/2$ factor changed to a larger $2\pi/3$.

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Intersecting the given paraboloid and the plane $z = \bar{z}$, where $\bar{z} \in [0,h]$, we obtain the ellipse

$$\frac{x^2}{\left(a \sqrt{1 - \frac{\bar{z}}{h}}\right)^2} + \frac{y^2}{\left( b \sqrt{1 - \frac{\bar{z}}{h}}\right)^2} = 1$$

It is known that the area of an ellipse is $\pi$ times the product of the lengths of the semi-major and semi-minor axes. Hence, an infinitesimally thin "slice" of the paraboloid at "height" $z \in [0,h]$ has the following infinitesimal volume

$$\mathrm d V = \pi a b \left( 1 - \frac{z}{h} \right) \mathrm d z$$

and, integrating over $[0,h]$, we obtain

$$V = \frac{\pi a b h}{2}$$

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  • $\begingroup$ No need to use calculus to compute the integral. What is the area of a triangle? $\endgroup$ – Rodrigo de Azevedo Apr 13 at 18:55

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