Calculate the volume of revolution generated by the function, f(x) = (3^x)√x, for the domain x = [0, 1]

First we recall the general formula for the volume of revolution: V = π ∫ [f(x)]² dx Substituting our function, this gives: V = π ∫₀¹ (3^x)² x dx We could write (3^x)² as 3^(2x), or we could notice that this is equal to 9^x. This avoids unnecessarily over-complicating our integration which could lead to mistakes. So we have: V = π ∫₀¹ (9^x) x dx Given that we have two different functions of x under the integral, we must use the integration by parts method. Deciding which element to differentiate and which to integrate is made easier when we realise that: ∫ (9^x) dx = (9^x)/ln(9) = (9^x)/2ln(3) To prove this, let y = a^x and take the natural logarithm of both sides: ln(y) = ln(a^x) Recall that one of our rules for logarithms is that ln(a^b) = bln(a): ln(y) = xln(a) Differentiate both sides with respect to x: y’/y = ln(a) dy/dx = yln(a) = (a^x)ln(a) Now multiply by dx, divide by ln(a) and integrate both sides: ∫ (1/ln(a)) dy = ∫ a^x dx Therefore ∫ a^x dx = y/ln(a) = (a^x)/ln(a) As such if we differentiate x and integrate (9^x), we should be left with an integral we can immediately solve given that d/dx[x] = 1. u = x, v’ = 9^x u’ = 1, v = (9^x)/2ln(3) Recall that the IBP formula is: I = vu - ∫ vu’ dx V = π [[ (9^x)x/2ln(3) ]₀¹ - (1/2ln(3)) ∫₀¹ 9^x dx ] Note that we have simply taken a constant out of the integral to simplify it. When evaluating the first term, we only need to consider when x =1 as the second term goes to 0 due to the x in the numerator. V = π [ 9/2ln(3) – (1/2ln(3)) [ (9^x)/2ln(3) ]₀¹] V = π [ 9/2ln(3) – (1/2ln(3)) [ 9/2ln(3) – 1/2ln(3) ]] Notice that we can factorise out 1/2ln(3) and there is a common denominator in the far-right bracket allowing us to sum the numerators, leaving 8/2ln(3). Cancel the 8 with the 2 and we’re left with: V = π/2ln(3) [ 9 - 4/ln(3) ]