The area between the y-axis and y= tan x for 0≤ x ≤ pi/4 is rotated about the y-axis.?
The area between the y-axis and y= tan x for 0≤ x ≤ pi/4 is rotated about the y-axis.
a) What is the volume of the resulting solid?
b) Suppose the solid is a foam with varying density given by 2/( pi arctan (y)) grams per cubic centimeter, where x and y are given in centimeters. Even though the foam has infinite density y = 0, the total mass is still finite. What is the total mass of the foam?
**thank you will vote best answer!!**
the answers are, a) 0.770574 and b) 0.877649... I need to know how they got the answers.. thank you!
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When x=π/4, y = tan(x) = 1, then the cap of the volume is the circle or radius 1 with center on the y axis, and located at the plane y=1.
The volume can be decomposed into cylindrical shells, with radii x and x+dx and height 1-tan(x). These shells have volume dV = (1-tan(x))π[(x+dx)² - x²] = 2π(1-tan(x))xdx.
The total volume is
V = ∫(from x=0 to π/4) 2πx(1-tan(x))dx =~ 0.770574 (numerically, it cannot be written as a combination of elementary functions).
Another way, integrating disk over y:
The solid is decomposed into disks or radius x = arctan(y) and height dy:
V = ∫(from y=0 to 1) π arctan²(y) dy =~ 0.770574 (again a numerical result, but it can be written in terms of an elliptic integral, or in terms of Catalan's constant...).
b)
M = ∫(from y=0 to 1) π arctan²(y) (2/(π arctan(y))) dy =
= ∫(from y=0 to 1) 2 arctan(y) dy = π/2 - ln(2) =~ 0.877649