Anyone know this ♀️ ??
1 answer:
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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.