Standar form of 1.204751131
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(i) Given that
when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then
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
Compute the integral:
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.
Multiply the equation:
The solution set is the same, because multiplying both sides of an equation by a non-zero number doesn't change the solution set. In fact, if you rewrite the equation as
Multiplying this by 3 (or whatever number, for all it matters) gives
Now, a product is zero if and only if at least one of the factor is zero. So, either or
Since the first is clearly impossible, the second one must be true, which is the original equation.