Answer: Provided the two functions, f(x) and g(x) we have to answer a to e parts:
(a) Graph of the region bounded by f and g and their points of intersection:
(b) Area of the region between the functions can be calculated by setting up an integral equation and then solving it, the integral is as follows:
[tex]\begin{gathered} \text{ Area = Area line - Area curve} \\ \\ \\ A=A_l-A_c \\ \\ \\ A=\int_{-1}^3g(y)dy-\int_{-1}^3f(y)dy \\ \\ \\ A=\int_{-1}^36dy-\int_{-1}^{-3}[(y-1)^2+2]dy=\int_{-1}^3[6-(y-1)^2-2[dy \\ \\ \\ A=\int_{-1}^3[6-(y-1)^2-2]dy\rightarrow(1) \end{gathered}[/tex]
The area is the value of the integral (1). Its value is as follows:
[tex]\begin{gathered} A=\int_{-1}^3(6-(y-1)^2-2)dy=10.6666 \\ \\ \\ A\approx10.667 \end{gathered}[/tex]
(c) Revolving the area between the functions around the y = -1.
When we include the line, the new graph looks as follows.
The new integral formulation, which would be used to calculate the volume of the new solid is as follows:
[tex]V=\int_2^6\pi[\sqrt{x-2}+2]^2dx-\int_2^6\pi[-\sqrt{x-2}+2]^2dx\rightarrow(2)[/tex]
The integral (2) represents the following:
Which is just the original f(y) shifted along the y-axis by 1 unit, the value of the integral (2) is as follows:
[tex]\begin{gathered} V=\int_2^6\pi[\sqrt{x-2}+2]^2dx-\int_2^6\pi[-\sqrt{x-2}+2]^2dx \\ \\ \\ \\ V=\pi\int_2^6([\sqrt{x-2}+2]^2-[-\sqrt{x-2}+2]^2)dx=\frac{32}{3}\pi \\ \\ \\ \\ V=\frac{32}{3}\pi \end{gathered}[/tex]
