1) Hello, please can you assist me with this urgently

Question

Find the area of the surface of the cap, cut from the

hemisphere x ² + y² + z² = 5, z ≥ 0, by the cylinder x² + y² = 1.

Answer 1

Let

R = 5 be the radius of the hemisphere,

r = 1 be the radius of the cylinder.

Imagine that the hemisphere is cut by many vertical cuts into many thin wedges.

Each wedge is identified by its angle θ from any vertical plane, for example, from the plane y=0.

The angle θ ranges between 0 and 2π.

The angle between two neighboring wedges is a diminishingly small dθ.

Secondly, the hemisphere is cut by many concentric vertical cylinders around the z-axis.

Each cylinder is identified by its radius ρ.

The radius ρ ranges between 0 and r.

The difference in radius between two neighboring cylinders is a diminishingly small dρ.

The vertical wedges and the cylinders divide the hemisphere into tiny shapes, which we can

approximate by rectangles.

For each of the small rectangle, one pair of sides is along two neighboring cylinders.

This pair of sides is horizontal, i.e., parallel to the x-y-plane.

For a rectangle located at radius ρ, this pair of sides has length ρdθ.

For each rectangle, the other pair of sides is along two wedges.

Let ds be the length of this pair of sides; we need to calculate ds in terms of dρ.

I recommend drawing a picture of such a vertical cut.

It contains an arc with radius R.

On that arc there is a point P, where our rectangle is located.

This point P is some distance ρ from the z-axis.

Draw a line (of length R) connecting P with the origin.

That line forms some angle α with the z-axis.

This angle satisfies

sin(α) = ρ/R

The tangent at the point P, being perpendicular to the radius at P, forms the same angle α with the horizontal.

Consider a small right angle triangle at P whose one side is horizontal and has length dρ.

Its hypothenuse is along the tangent and has length ds.

In that triangle

cos(α) = dρ/ds.

Thus

ds = dρ/cos(α) = dρ/√(1-sin²(α)) = dρ/√(1-ρ²/R²)

To sum up the above result, we divided the surface A to be calculate into small rectangles

with dimensions ρdθ×dρ/√(1-ρ²/R²), where

ρ ranges between 0 and r, and θ ranges between 0 and 2π.

So now we just sum up (i.e., integrate) the areas of the rectangles as ρ and θ range between their bounds.

First calculate the indefinite integral

F(ρ) = ∫ρdρ/√(1-ρ²/R²) (1)

Use the substitution

u = 1-ρ²/R²

Then

du/dρ = -2ρ/R²

So

ρdρ = -R²du/2

The integral (1) can be rewritten as

F(ρ) = (-R²/2)∫du/√u = (-R²/2) 2√u = -R²√(1-ρ²/R²)

Thus the definite integral ∫ρdρ/√(1-ρ²/R²) between 0 and r is

F(r) - F(0) = (-R²√(1-r²/R²)) - (-R²√(1-0²/R²)) = R²(1 - √(1-r²/R²)).

And this constant needs to be integrated between 0 and 2π, which gives the area

A = 2πR²(1 - √(1-r²/R²)) as the final answer.

Note that 2πR² is the surface of the hemisphere.

It is being multiplied by a factor ≤ 1 --

the fraction of the hemisphere with radius R cut out by the cylinder of radius r.

The expression for the area A can be algebraically manipulated into

A = πr²(2/(1 + √(1-r²/R²)))

Note that πr² is the cross section of the cylinder of radius r.

It is being multiplied by a factor 1 ≤ 2/(1 + √(1-r²/R²)) ≤ 2

When r/R is very small then the factor is close to 1 and dome above the cylinder

has area very similar to cross section of the cylinder.

This is because the top of the dome is close to flat.

In contrast, when r is close to R then the factor is close to 2, and the area A

become close to the area of the hemisphere.