Question

A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring shaped solid that remains.

Answer 1

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

Volume of Ring = 21 unit^3

Answer 2

The volume of the ring-shaped solid that remains is 113.097

If we take a look at the sphere is bounded by the region;

$\mathbf{x^2+y^2+z^2 = r_2^2}$

$\mathbf{z =\sqrt{ r_2^2 -x^2-y^2}}$

Volume  $\mathbf{V =\iint_D\sqrt{ r_2^2 -x^2-y^2} \ dA}}$

where:

• D = region between circles of radius $\mathbf{r_1 \ and \ r_2}$ which is just half of the volume;

Thus, the volume remaining can be expressed as:

$\mathbf{V =2\iint_D\sqrt{ r_2^2 -x^2-y^2} \ dA}}$

Using polar coordinates, we know that:

$\mathbf{\Big \{ (r,0) \ \varepsilon \ D\Big| r_1<r<r_2, 0 < \theta <2 \pi \Big\}}$

∴

$\mathbf{V =2\int^{2}_{0} \int^{r_2}_{r_1} \sqrt{ r_2^2 -r^2(r)} \ dr \ d \theta }}$

$\mathbf{V =4 \pi \int^{r_2}_{r_1} r \sqrt{ r_2^2 -r^2} \ dr \ }}$

Assuming $\mathbf{ r_2^2 -r^2= U }}$;

• then, $\mathbf{ -2rdr= dU}}$

Taking the limits of the integration:

$\mathbf{\int^{r_2}_{r_1} = \int ^{0}_{r_2 -r_1^2}}$

∴

$\mathbf{Volume (V) = -2 \pi \int^0_{r_2^2-r_1^2} \ U^{1/2} \ du}}$

$\mathbf{\implies-2 \pi \Big [\dfrac{2}{3}U^{3/2} \Big ]^0_{r_2^2-r_1^2} }$

$\mathbf{\implies\dfrac{4 \pi}{3} (r_2^2-r_1^2)^{3/2} }$

From the question;

• r₂ = 5 cm
• r₁ = 4cm

∴

$\mathbf{\implies\dfrac{4 \pi}{3} (5^2-4^2)^{3/2} }$

$\mathbf{\implies\dfrac{4 \pi}{3} (\sqrt{9}^{3}) }$

$\mathbf{\implies\dfrac{4 \pi}{3} \times 27 }$

Volume (V) = 36 π

Volume (V) = 113.097

Therefore, we can conclude that the volume of the ring-shaped solid that remains is 113.097

Learn more about the volume of a sphere here:

https://brainly.com/question/16686115?referrer=searchResults