Let R be the region in the first quadrant bounded by the graphs of y= lnx, 1=x ,and 1=y as shown in the figure above.

Question

Let R be the region in the first quadrant bounded by the graphs of y= lnx, 1=x ,and 1=y as shown in the figure above.

(a) Find the area of R

(b) Region R is the base of a solid. For each y where 0 ≤ y ≤ 1, the cross-section of the solid taken perpendicular to the y-axis is a rectangle whose height is 4 times the length of its base in region R. Find the volume of this solid.

Answer 1

(a) f(x) = 1, g(x) = ln(x)

f(x) intersects with g(x) when 1 = ln(x), x = e

A = ∫₁^e 1-ln(x) dx = x-xln(x) + x = x(2-ln(x))|₁^e = e(2-1)-(2-0) = e-2

(b)

Volume V = ∫Ady

A = base×height = 4b×b = 4b²

y = ln(x) --> x = e^y

b = e^y-1

V = ∫₀¹4(e^y-1)²dy

= 4 ∫₀¹ e^2y-2e^y+1 dy

= 4[e^2y/2 -2e^y+y]₀¹= 4[e²/2-2e+1 - (1/2 -2 + 0)]

= 4(e²/2-2e + 5/2) = 4(e²-4e+5)/2 = 2e²-8e+10 ≈ 3.03 units³

Answer 2

Intersection lines y = lnx and y = 1; lnx = 1; x = e. And x = 1.

(a) Area A = ∫₁^e(1 - lnx)dx = (x - (xlnx - x))₁^e = (2x - xlnx)₁^e = (2e - elne) - (2) = e - 2 = 0.7182818285

Volume V = ∫₁^e 4(lnx - 1)²dx = 1.746254628.

This integral we used TI-84 we used fnInt function of Math menue