A hole of a radius of 1cm is pierced in a sphere of a 4cm radius. Calculate the volume of the remaining sphere.
I know that the volume of a sphere with an exterior radius is of:
V = pie(R^2 - r^2) dH
where R^2 is the exterior surface area and r^2 is the interior surface area. dH is the thickness.
for R and r, do I simply just plug in the values that are given to me?
Thank you.
The answer depends upon where the hole is drilled. Is it centered along a diameter of the sphere?
yes , the hole goes through the sphere, centered along the diameter
So far so good.
h^2 = R^2 - x^2
2h dh = -2x dx
or,
dh = -x/h dx
so the integral becomes
pi (R^2-r^2) (-x/sqrt(R^2 - x^2)) dx
= -pi (R^2-r^2) x/sqrt(R^2-x^2) dx
now let u = R^2 - x^2
du = -2x dx
and the integral is
pi/2 (R^2-r^2) u^(-1/2) du
If I've done things right,
v = 4/3 pi (R^2-r^2)^(3/2)
makes sense, since if r=0, you have the whole sphere, and if r=R, there's nothing.
I see how it makes sense now, thanks a lot Steve!
Yes, to calculate the volume of the remaining sphere, you can use the formula you mentioned:
V = π(R^2 - r^2)dH
To plug in the values, let's assign the given values:
R = 4 cm (exterior radius of the original sphere)
r = 1 cm (radius of the hole)
We need to calculate dH, which represents the thickness of the remaining shell of the sphere.
To calculate dH, we can subtract the radius of the hole from the exterior radius:
dH = R - r
= 4 cm - 1 cm
= 3 cm
Now, we have all the required values to calculate the volume of the remaining sphere:
V = π(R^2 - r^2)dH
= π((4 cm)^2 - (1 cm)^2) * 3 cm
= π(16 cm^2 - 1 cm^2) * 3 cm
= π(15 cm^2) * 3 cm
= 45π cm^3
Therefore, the volume of the remaining sphere is 45π cm^3.