Volume integral, calculating mass

Nov 2019
22
1
UK
Can anyone help please?

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
Sep 2015
2,958
1,673
USA
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.
 
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Nov 2019
22
1
UK
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!

May I ask why its
4api and not 4pi and then multiplying inside the integral by r^2?

Also, for the mass of the cylinder portion I have a^2/2*a + a

For the mass of the dome I have (7/3)*pi*a^2 - a*pi

Is this correct? and do I just add them together?

Thank you