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.