CALCULUS HELP NEEDED PLEASE!!!!1.) Find the area or the region bounded by the curves y = x^3 and y = 9x.a. 0b. 10.13c. 40.50d. 20.252.)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. 1.219b. 3.830c. 1.786d. 5.612

Accepted Solution

1. The given curves intersect one another three times:[tex]x^3=9x\implies x(x^2-9)=0\implies x=0,\pm3[/tex]The area of the bounded region is[tex]\displaystyle\int_{-3}^3|x^3-9x|\,\mathrm dx[/tex][tex]x^3-9x[/tex] is odd, but the absolute value makes it even. More formally,[tex]|(-x)^3-9(-x)|=|-x^3+9x|=|x^3-9x|[/tex]which means the integral is equivalent to[tex]\displaystyle2\int_0^3|x^3-9x|\,\mathrm dx[/tex]For [tex]0\le x\le 3[/tex], the definition of absolute value tells us that[tex]|x^3-9x|=9x-x^3[/tex]so the integral evaluates to[tex]\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[/tex]2. Using the disk method, the volume is given by the integral[tex]\displaystyle\pi\int_0^\pi\sin^2(\sin x)\,\mathrm dx[/tex]Use a calculator to get the result 1.219.