# Volume integral, calculating mass

#### Hamlings

A bullet-shaped object is formed from a cylinder with a curved top. The cylinder has radius a and is situated with its base on the (x, y)-plane, with the origin located at the centre of the base. The curved top of the bullet is given by the formula z = 3a − x^ 2/a − y ^2/a
and the density is given by D( (sqrt(x^2+y^2)/a) +1)
where a and D are constants.
Calculate the mass of the object.

#### romsek

Math Team
Well first thing to do is determine where the cylinder ends and the dome starts.

$z = 3a -\dfrac 1 a (x^2 + y^2) = 3a - \dfrac{r^2}{a} = 3a - a = 2a$

So for the cylindrical portion we simply have

$m_{cyl} = \displaystyle \int_0^{2a} \int_0^{2\pi} \int_0^a \dfrac{r}{a} + 1~dr~d\theta~dz = 4a\pi \displaystyle \int_0^a \dfrac{r}{a}+1~dr$

The dome portion starts at $2a$ and has a maximum at $3a$

$m_{dome} = 2 \pi \displaystyle \int_0^a \int_{2a}^{3a-\frac{r^2}{a}} \dfrac r a + 1 ~dz~dr$

I leave you to evaluate those integrals and finish up.

topsquark

#### Hamlings

Well first thing to do is determine where the cylinder ends and the dome starts.

$z = 3a -\dfrac 1 a (x^2 + y^2) = 3a - \dfrac{r^2}{a} = 3a - a = 2a$

So for the cylindrical portion we simply have

$m_{cyl} = \displaystyle \int_0^{2a} \int_0^{2\pi} \int_0^a \dfrac{r}{a} + 1~dr~d\theta~dz = 4a\pi \displaystyle \int_0^a \dfrac{r}{a}+1~dr$

The dome portion starts at $2a$ and has a maximum at $3a$

$m_{dome} = 2 \pi \displaystyle \int_0^a \int_{2a}^{3a-\frac{r^2}{a}} \dfrac r a + 1 ~dz~dr$

I leave you to evaluate those integrals and finish up.
Thank you!