Two functions, y1 & y2, are given by y1=x^2+16x+4; y2=2(3x+2). Find analytically the volume of the solid created by revolving the area between the two curves by 2pi radians around the x-axis. N.B. y2>y1 on the interval between the points of intersection.

Setting the two equations equal to one another to find the intersection boundaries:x^2+16x+4=6x+4; x^2+16x+4-6x-4=0; x^2+10x=0;x(x+10)=0; x=0, x=-10. Setting up the volume of revolution (N.B. Due to typing restrictions, the boundaries will be applied at the end): V=|piInt((y^2)dx)|, where y=y2-y1, as specified by the question. Doing the integration separately to simplify the calculations: I=Int((y^2)dx)=Int(-(x^2+10x)^2dx)=Int(-(x^4+20x^3+10x^2)dx)=-(x^5/5+5x^4+100x^3/3) Hence, substituting the result above into the original volume equation, and taking common fractions: V=|-pi/15(3x^5+75x^4+500x^3)| Applying the boundaries: V=|-pi/15*(-1)(-310^5+7510^4-50010^3)|=|-0.5pi/1510^5|=|-pi/3010^5|. Therefore, the Volume of the generated solid is pi/30*10^5 units cubed.