Tough Integration Qn on Volume, R(x)= (10-x)(x^2)/10?

Question

The profile of an experimental propeller is given by R(x)= (10-x)(x^2)/10 for 0
1) What volume, V(t), of water is swept out by the leading portion of the blade in one revolution? (Enter an expression in terms of t)

2) What is the total volume of water swept out by the blades in one revolution?

Answer 1

Note from the form of R(x) that R(x) is positive on (0, 10) and zero at 0, 10. So the volume integral is fairly simple.

1. V(t) = π int(0 to t) (R(x)^2) dx
= π int(0 to t) [(100 - 20x + x^2) (x^4) / 100] dx
= π [x^5 / 5 - (1/5) x^6 / 6 + (1/100) x^7 / 7] [0 to t]
= π (t^5 / 5 - t^6 / 30 + t^7 / 700)

2. This is just V(10), i.e.
π (10^5 / 5 - 10^6 / 30 + 10^7 / 700)
= 952.4 cm^3 to 4 s.f.