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
Answers
Answer:
The volume common to two spheres is [tex]V=\frac{27 \pi r^3}{32}[/tex]
Step-by-step explanation:
Let r be the radius of each sphere
We are given that distance between their centers is [tex]\frac{r}{2}[/tex]
Let [tex](\frac{-r}{4},0)[/tex] be the center of sphere 1 and [tex](\frac{r}{4},0)[/tex]be the center of sphere 2 .
Using the disk points
a=0 and b =[tex]\frac{3r}{4}[/tex]
[tex](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}[/tex]
Hence the volume common to two spheres is [tex]V=\frac{27 \pi r^3}{32}[/tex]
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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Step-by-step explanation:
Let's break this down...
9.03+10.71 is 19.74 and subtracted by 6.8 it will be 12.94 not 13.94 so it is not correct
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Hope this helps!
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Hi there!
We are given that the slope of a line is -1 and the y intercept of that same line is 8
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