How do you do this problem?

Step-by-step explanation:

My work is in the attachment, but the idea is to integrate the area of a circle along some small distance to get the volume of a shape whose sides follow the graph (r isn't constant).

Comment with any questions.

V = 8π²/3 − 2π√3

Step-by-step explanation:

Start by drawing the region.  Evaluate the function at the endpoints.

f(x) = y = 1 + sec(x)

f(-π/3) = 1 + sec(-π/3) = 3

f(π/3) = 1 + sec(π/3) = 3

Therefore, the curve is the one shown in the left side options.  We want the region between the curve and y=3, so the correct sketch is the lower left.

When we revolve this region about y=1, we get a hollow cylinder on its side (again, the answer in the lower left).

Slicing vertically, we get washers with thickness dx, outside radius of 3 − 1 = 2, and inside radius of 1 + sec(x) − 1 = sec(x).  The volume of the washer is:

dV = π (R² − r²) t

dV = π (2² − sec²(x)) dx

dV = π (4 − sec²(x)) dx

The total volume is the sum of all the washers from x=-π/3 to x=π/3.

V = ∫ dV

V = ∫ π (4 − sec²(x)) dx

V = π (4x − tan(x))

Evaluate from x=-π/3 to x=π/3.

V = π (4(π/3) − tan(π/3)) − π (4(-π/3) − tan(-π/3))

V = π (4π/3 − √3) − π (-4π/3 + √3)

V = 4π²/3 − π√3 + 4π²/3 − π√3

V = 8π²/3 − 2π√3

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.