Question

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

Answer 1

Answer:

The volume common to two spheres is $V=\frac{27 \pi r^3}{32}$

Step-by-step explanation:

Let r be the radius of each sphere

We are given that distance between their centers is $\frac{r}{2}$

Let $(\frac{-r}{4},0)$ be the center of sphere 1 and $(\frac{r}{4},0)$be the center of sphere 2 .

Using the disk points

a=0 and b =$\frac{3r}{4}$

$(x+(\frac{r}{4}))^2+y^2=r^2\\V=2 \int_{0}^{\frac{3r}{4}} \pi y^2 dx\\V=2 \int_{0}^{\frac{3r}{4}} \pi [r^2-(x+(\frac{r}{4}))^2]dx\\V=\frac{27 \pi r^3}{32}$

Hence the volume common to two spheres is $V=\frac{27 \pi r^3}{32}$

Answer 2

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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