Question

Let R be the region bound by the equations y = 2 + cos(x) and y = csc(x) in the first quadrant on theinterval 0 ≤ x < π.

Answer 1

Answer:

b.

$V = \pi \cdot \int\limits^a_b {\left([f(x)]^2 - [g(x)]^2} \right) \, dx$

(c)

$V = \pi \cdot \int\limits^3_1 {\left([arcos(y - 2)]^2 - [arcsine(x)]^2 - (-1)^2} \right) \, dx$

Step-by-step explanation:

b. The volume of solid formed is given by the washers formula as follows;

$V = \pi \cdot \int\limits^a_b {\left([f(x)]^2 - [g(x)]^2} \right) \, dx$

Therefore, we have, the integral expression whose solution is the volume formed by rotating 'R', about the equations y = 2 + cos(x) and y = csc(x) in the first quadrant on the interval, 0 ≤ x ≤ π, V is given as follows;

$V = \pi \cdot \int\limits^\pi_0 {\left([2 + cox(x)]^2 - [csc(x)]^2} \right) \, dx$

(c)  We have;

x = arcos(y - 2), x = arcsin(1/y)

At x = 0, y = 2 + cos(0) = 3

csc(0) = ∞

At x = π, y = 2 + cos(π) = 2 + -1 = 1

csc(π) = ∞

Therefore, we get;

$V = \pi \cdot \int\limits^3_1 {\left([arcos(y - 2)]^2 - [arcsine(x)]^2 - (-1)^2} \right) \, dx$