Question:
Let R be the region enclosed by the graph of f(x) = x4 – 2.3x3 + 4 and the horizontal line y = 4, as shown in the figure above.
(a) Find the volume of the solid generated when R is rotated about the horizontal line y = -2.
(b) Region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is an isosceles right triangle with a leg in R. Find the volume of the solid.
(c) The vertical line x = k divides R into two regions with equal areas. Write, but do not solve, an equation involving integral expressions whose solution gives the value k.
Answer/Explanation
Ans:
(a)
\(f(x)= 4| _{x = 0 or 2.3} \) \(\pi \int_{0}^{2.3}\left ( (2+4)^{2}-(2+f(x))^{2} \right )dx = 98.868 units^{3}\)
\(V = \pi \left ( R^{2}-r^{2} \right )\)
(b)
\(A = \frac{1}{2}bh\)
\(A = \frac{1}{2}b^{2}\)
b = 4 – f(x)
\(\frac{1}{2}\int_{0}^{2.3}(4-p(x))^{2}dx= 3.574 units^{3}\)
(c)
\(\int_{0}^{k}(4-f(x))dx= \frac{1}{2}\int_{0}^{2.3}\left ( 4-f(x) \right )dx\)