1. Answer to the vase problem
Volume of vase = πZ24
0
g2(y)dy.
Add a row to the table to record the values of g2(y)
y0 8.0 16.0 24.0
g(y) 7.0 4.0 5.0 6.0
g2(y) 49.0 16.0 25.0 36.0
The trapezoidal formula is
Tn=h
2(y0+ 2y1+· · · + 2yn−1+yn),
h=b−a
n
where that are nintervals and n+ 1 points, on the interval of integration [a, b].
In our case n= 3, a = 0, b = 24, h= 8 and the approximation of R24
0g2(y)dy is
Tn=8
2(49 + 2 ×16 + 2 ×25 + 36)
= 668
Thus the approximation to πR24
0g2(y)dy is 668π.