Question

Let R1 be the region in the 1st quadrant bounded by y=x, the y-axis,&the line y=1. Let R2 be the region in the 1st quadrant bounded by y=x2&y=x. Let R3 be the region in the 1st quadrant bounded by y=x2,the x-axis,&the line x=1.Write an integral expression for the volume of the slid described. The volume of the solid generated by revolving R1 about the line y=-1.

Answer 1

\[\pi\int_{0}^{1}1^2-x^2dx\]

Answer 2

want some explanation?

Answer 3

that wud b great

Answer 4

actually I messed up..\[\pi\int_{0}^{1}2^2-(x+1)^2dx\]look at the drawing:notice that the outer radius of our discs will be a radius 2 from the line y=-1
and the inner disc will be a distance f(x)+1=x+1 from the line y=1
The integration is from x=0 to x=1 because x for the shaded area R1

The disc method formula for a hollowed shape is\[\pi\int_{0}^{1}r_1^2-r_2^2dx=\pi\int_{0}^{1}2^2-(x+1)^2dx\]