Question

CALCULUS HELP NEEDED PLEASE!!!!

1.) Find the area or the region bounded by the curves y = x^3 and y = 9x.
a. 0
b. 10.13
c. 40.50
d. 20.25

2.) The region in the first quadrant bounded by the x-axis, the line x = π, and the curve y = sin(sin(x)) is rotated about the x-axis. What is the volume of the generated solid?
a. 1.219
b. 3.830
c. 1.786
d. 5.612

Answer 1

1. The given curves intersect one another three times:

`x^3=9x\implies x(x^2-9)=0\implies x=0,\pm3`

The area of the bounded region is

`\displaystyle\int_(-3)^3|x^3-9x|\,\mathrm dx`

`x^3-9x` is odd, but the absolute value makes it even. More formally,

`|(-x)^3-9(-x)|=|-x^3+9x|=|x^3-9x|`

which means the integral is equivalent to

`\displaystyle2\int_0^3|x^3-9x|\,\mathrm dx`

For
`0\le x\le 3`, the definition of absolute value tells us that

`|x^3-9x|=9x-x^3`

so the integral evaluates to

`\displaystyle2\int_0^3(9x-x^3)\,\mathrm dx=\left(9x^2-\frac{x^4}2\right)\bigg|_(x=0)^(x=3)=\frac{81}2=40.5`

2. Using the disk method, the volume is given by the integral

`\displaystyle\pi\int_0^\pi\sin^2(\sin x)\,\mathrm dx`

Use a calculator to get the result 1.219.