Calculate the volume enclosed by the paraboloid $f(x,y)=1-\frac{1}{2}x^2-\frac{1}{2}y^2$

Question

Calculate the volume enclosed by the paraboloid $f(x,y)=1-\frac{1}{2}x^2-\frac{1}{2}y^2$ and the plane $z=0$, when $f(x,y)$ is defined in $D=[0,1]\times [0,1]$.

I used polar coordinates and I had the following integral,

$$\int _{0}^{\frac{\pi}{2}} \int _{0}^{1} \left(1-\frac{1}{2}r^2\right)\,dr\,d\theta =\dfrac{3\pi}{16}$$

Is it right?

Answer 1

With Cartesian coordinates, volume is $$\int _{0}^{1}\int _{0}^{1}\int _{0}^{1-\frac12 x^2-\frac12 y^2}dzdydx=0.7$$

Answer 2

$$\int\limits_{x=0}^1 \int\limits_{y=0}^1 \int\limits_{z=0}^{1 - x^2/2 - y^2/2} 1\ dx\ dy\ dz = \frac{2}{3}$$

Answer 3

In this case cartesian coordinates are convenient.

Using polar cordinates the correct set up should be

$$\int_0^{\frac \pi 2} d\theta \int_0^1 dz \int_0^{\sqrt{2-2z}}rdr-2\int_0^{\frac \pi 4}\int_{\frac{1}{\cos \theta}}^\sqrt 2 d\theta\int_0^{1-\frac {r^2}2}dz=\frac \pi 4-2\left(\frac \pi 8-\frac1{3}\right)=\frac23$$

• first integral
• second integral