Find the volume common to two spheres, each with radius r, if the distance between their centers is r/2. [Hint: It may be helpful to look at the previous question and see how it relates to this one. ] intersectingspheres

In attached diagram, we can see 2 circumferences with radius r, and separated centers by r/2.

The solution is:

V = (11/12)×π×r³

Let´s call circumferences  1 and 2, by symmetry, rotating area A will produce a volume V₁ identical a V₂, Obtained by rotating area B ( both around x-axis), then the whole volume V will be:

V = 2× V₁

V₁ = ∫π×y²×dx         (1)

Now

( x - r/2)²  +  y² = r²       the equation of circumference 1

y² = r² - ( x - r/2)²

Plugging this value in equation (1)

V₁ = ∫π×[  r² - ( x - r/2)²]×dx        with integrations limits    0 ≤ x ≤ r/2

V₁ = π×∫ ( r² - x² + (r/2)² - r×x )×dx

V₁ = π× [  r²×x - x³/3 + (r/2)²×x - (1/2) ×r×x²]    evaluate between 0 and r/2

V₁ = π× [(5/4)×r²×x - x³/3 - (1/2) ×r×x²]

V₁ =  π× [(5/4)×r² × ( r/2 - 0 ) - (1/3)×(r/2)³ -  (1/2) ×r×(r/2)²]

V₁ =  π× [ (5/8)×r³ - r³/24 - r³/8]

V₁ =  π× (11/24)×r³

Then

V = 2× V₁

V = 2×π×11/24)×r³

V = (11/12)×π×r³

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The volume common to two spheres is

Step-by-step explanation:

Let r be the radius of each sphere

We are given that distance between their centers is

Let be the center of sphere 1 and be the center of sphere 2 .

Using the disk points

a=0 and b =

Hence the volume common to two spheres is

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.

2 x t divided by n

step-by-step explanation: