Which of the following integrals computes the volume of the

Solution

If we rotate this area around the x axis, we get a more or less conically-shaped solid. If we cut this solid into a whole lot of slices transverse to the x axis, find the volume of each slice, then add them all up, we will have the volume of the solid. If we slice the solid into slices of thickness dx, then each slice that lies at point x will have an area of py² = p ln²(x) since y = ln(x) at that point. Therefore, the volume of the slice is p ln²(x) dx Adding these up (i.e. integrating them) between x=1 and x=e, we get p?{1 to e} ln²(x) dx The integral is easily done by parts*, and resolves to x[ ln²(x) - 2ln(x) + 2 ], so the result is p x[ln²(x) - 2ln(x) + 2] | {between e and 1} = p [e( ln²(e) - 2ln(e) + 2 ) - 1( ln²(1) - 2ln(1) + 2 )] = p [e( 1 - 2 + 2 ) - 1( 0 - 0 + 2 ) ] = p (e - 2) * To integrate this by parts, use u = ln²(x) and dv = dx. Then du = 2 ln(x)/x and v = x, and since u dv = uv - v du, ? ln²(x) dx = x ln²(x) - ?2x ln(x)/x dx = x ln²(x) - 2?ln(x) dx Now do this second integral by parts, with u = ln(x) and dv = dx, so that du = dx/x and v = x: x ln²(x) - 2?ln(x) dx = x ln²(x) - 2[ x ln(x) - ?x/x dx ] = x ln²(x) - 2x ln(x) + 2x = x[ ln²(x) - 2 ln(x) + 2 ]