# Volume of a truncated paraboloid

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!

## 3 Answers

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 $$V=\int \mathrm{d}V=\int_{-a}^{a}\mathrm{d}x\left(\int_{\hat{y}_{-}(x)}^{\hat{y}_{+}(x)} \mathrm{d}y\left(\int_0^{\hat{z}(x,y)}\mathrm{d}z\right)\right)$$

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 $$2\pi/3$$.

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

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

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