(i) Given that
[tex]V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr[/tex]
when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then
[tex]V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}[/tex]
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
[tex]\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr[/tex]
Compute the integral:
[tex]V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr[/tex]
[tex]V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr[/tex]
[tex]V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R[/tex]
[tex]V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right) [/tex]
[tex]V(R) = \displaystyle \boxed{\frac{\pi KR^4}2} [/tex]
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.
