(i) Given that
![V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr](https://tex.z-dn.net/?f=V%28R%29%20%3D%20%5Cdisplaystyle%20%5Cint_0%5ER%202%5Cpi%20K%28R%5E2r-r%5E3%29%20%5C%2C%20dr)
when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then
![V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}](https://tex.z-dn.net/?f=V%280.30%29%20%3D%20%5Cdisplaystyle%20%5Cint_0%5E%7B0.30%7D%202%5Cpi%20%5Cleft%280.30-3.33r%5E2%5Cright%29r%20%5C%2C%20dr%20%5Capprox%20%5Cboxed%7B0.0425%7D)
and this is a volume so it must be reported with units of cm³.
In Mathematica, you can first define the velocity function with
v[r_] := 0.30 - 3.33r^2
and additionally define the volume function with
V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]
Then get the desired volume by running V[0.30].
(ii) In full, the volume function is
![\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5ER%202%5Cpi%20K%28R%5E2-r%5E2%29r%20%5C%2C%20dr)
Compute the integral:
![V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr](https://tex.z-dn.net/?f=V%28R%29%20%3D%20%5Cdisplaystyle%20%5Cint_0%5ER%202%5Cpi%20K%28R%5E2-r%5E2%29r%20%5C%2C%20dr)
![V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr](https://tex.z-dn.net/?f=V%28R%29%20%3D%20%5Cdisplaystyle%202%5Cpi%20K%20%5Cint_0%5ER%20%28R%5E2r-r%5E3%29%20%5C%2C%20dr)
![V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R](https://tex.z-dn.net/?f=V%28R%29%20%3D%20%5Cdisplaystyle%202%5Cpi%20K%20%5Cleft%28%5Cfrac12%20R%5E2r%5E2%20-%20%5Cfrac14%20r%5E4%5Cright%29%5Cbigg_0%5ER)
![V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)](https://tex.z-dn.net/?f=V%28R%29%20%3D%20%5Cdisplaystyle%202%5Cpi%20K%20%5Cleft%28%5Cfrac%7BR%5E4%7D2-%20%5Cfrac%7BR%5E4%7D4%5Cright%29%20)
![V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}](https://tex.z-dn.net/?f=V%28R%29%20%3D%20%5Cdisplaystyle%20%5Cboxed%7B%5Cfrac%7B%5Cpi%20KR%5E4%7D2%7D%20)
In M, redefine the velocity function as
v[r_] := k*(R^2 - r^2)
(you can't use capital K because it's reserved for a built-in function)
Then run
Integrate[2 Pi v[r] r, {r, 0, R}]
This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with
Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]
which ensures that R is positive, and moreover a real number.