- Read the radius from the diagram. - Use the sphere volume formula. - Keep \(\pi\) in the exact answer.

From the diagram, the radius is \(3\,\text{cm}\). Using \(V=\frac{4}{3}\pi r^3\), \(V=\frac{4}{3}\pi\cdot3^3=36\pi\,\text{cm}^3\).

\(36\pi\,\text{cm}^3\)

- Think about the standard volume relationship for a sphere rather than a surface-area formula. - Check that the radius appears with the power expected for a three-dimensional measurement.

1. The volume formula for a sphere is \(V=\frac{4}{3}\pi r^3\).

\(V=\frac{4}{3}\pi r^3\)

- Does the marked segment stop at the center or continue to the other side? - How do a radius and diameter of the same circle compare?

1. The marked segment runs all the way across the sphere through its center, so it shows the diameter. 2. The radius goes only from the center to the surface, so \(r=10\div2=5\,\text{cm}\).

The marked segment crosses the whole sphere through its center, so it is the diameter, \(10\,\text{cm}\). The radius is half of that, \(5\,\text{cm}\).

- What kind of units count the space inside a solid rather than its surface? - If the radius were \(2\,\text{cm}\), what units would accompany its third power?

1. Cubing a length multiplies three centimeter measurements, so \(r^3\) has units \(\text{cm}^3\). 2. The number \(\frac{4}{3}\pi\) does not change the units. 3. The volume is therefore measured in cubic centimeters.

Cubic centimeters, \(\text{cm}^3\), because the radius is cubed.

- Identify which part of the formula changes. - The radius appears three times in \(r^3\). What happens to each factor when the radius doubles?

1. Sphere volume is \(\frac{4}{3}\pi r^3\). 2. Replacing \(r\) by \(2r\) changes \(r^3\) to \((2r)^3=8r^3\). 3. The volume is multiplied by \(8\).

The volume is multiplied by \(8\), because \((2r)^3=8r^3\).

- Convert the diameter to a radius first. - The radius is cubed in the sphere volume formula. - Leave \(\pi\) in the exact answer.

The radius is \(10\div2=5\,\text{cm}\). Then \(V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\cdot5^3=\frac{500\pi}{3}\,\text{cm}^3\).

\(\frac{500\pi}{3}\,\text{cm}^3\)

- Check whether the marked segment is the full width of the sphere. - Use half the diameter as the radius. - Cube the radius in the sphere-volume formula.

1. The displayed diameter is \(12\,\text{cm}\), so the radius is \(6\,\text{cm}\). 2. Substitute into the sphere formula: \(V=\frac{4}{3}\pi\cdot6^3=288\pi\,\text{cm}^3\).

\(288\pi\,\text{cm}^3\).

- Find the volume of the corresponding full sphere first. - Use the fact that a hemisphere is one of two equal halves of a sphere.

1. A full sphere of radius \(6\,\text{cm}\) has volume \(V=\frac{4}{3}\pi\cdot6^3=288\pi\,\text{cm}^3\). 2. A hemisphere has half that volume, so \(V=144\pi\,\text{cm}^3\).

\(144\pi\,\text{cm}^3\)

- Does the marked measurement stop at the center or continue to the opposite side? - What distance from the center does the formula call \(r\)? - How can you check whether the final volume has the correct units?

1. The diagram gives diameter \(6\,\text{cm}\), so the radius is \(3\,\text{cm}\). 2. Substitute: \(V=\frac{4}{3}\pi\cdot3^3\). 3. Simplify to \(V=36\pi\,\text{cm}^3\).

\(36\pi\,\text{cm}^3\).

- Substitute the known volume into the sphere formula. - Isolate \(r^3\) before taking a cube root. - Once the radius is known, relate it to the diameter.

Substitute into \(V=\frac{4}{3}\pi r^3\): \(288\pi=\frac{4}{3}\pi r^3\). Cancel \(\pi\) and multiply by \(\frac{3}{4}\): \(r^3=216\). Thus \(r=6\,\text{cm}\), so the diameter is \(12\,\text{cm}\).

Radius: \(6\,\text{cm}\); diameter: \(12\,\text{cm}\)

- Keep the original radius as a variable. - Replace the new radius by \(3r\) in the sphere formula. - Pay attention to the exponent on the radius.

Let sphere A have radius \(r\), so \(V_A=\frac{4}{3}\pi r^3\). Sphere B has radius \(3r\), so \(V_B=\frac{4}{3}\pi(3r)^3=27\left(\frac{4}{3}\pi r^3\right)=27V_A\). Therefore, tripling the radius multiplies the volume by \(27\).

Sphere B has \(27\) times the volume of sphere A.

- Read the sphere's radius from the diagram. - Cube the radius before multiplying by the remaining factors. - Round only the final volume.

From the diagram, the radius is \(4.5\,\text{cm}\). Then \(V=\frac{4}{3}\pi(4.5)^3=121.5\pi\,\text{cm}^3\approx381.7\,\text{cm}^3\).

About \(381.7\,\text{cm}^3\)

- Represent the new radius as a scale factor times the original radius. - Track how that scale factor is affected by the exponent in the sphere volume formula. - Work backward from the given volume factor.

1. Sphere volume is proportional to the cube of the radius, so the volume scale factor is \(k^3\). 2. Since the volume factor is \(8\), \(k^3=8\). 3. Therefore, \(k=2\).

\(k=2\)

- Which length does the sphere formula require? - Does a measurement all the way across a ball match that length? - Check the units in your final volume.

1. The \(8\,\text{cm}\) distance across the ball is its diameter, not its radius. 2. The radius is \(4\,\text{cm}\), so the correct formula is \(V=\frac{4}{3}\pi\cdot4^3\). 3. The correct volume is \(\frac{256}{3}\pi\,\text{cm}^3\).

Leo used the diameter as the radius. The correct volume is \(\frac{256}{3}\pi\,\text{cm}^3\).

- Read the dimensions from the two solids before computing either volume. - Use the sphere and cylinder volume formulas separately. - For part b, set the two general formulas equal. - Cancel only factors that are nonzero for a genuine solid.

1. From the diagram, both solids have radius \(3\,\text{cm}\), and the cylinder has height \(4\,\text{cm}\). The sphere volume is \(\frac{4}{3}\pi\cdot3^3=36\pi\,\text{cm}^3\). The cylinder volume is \(\pi\cdot3^2\cdot4=36\pi\,\text{cm}^3\), so they are equal. 2. Set the general volumes equal: \(\frac{4}{3}\pi r^3=\pi r^2h\). For \(r>0\), cancel \(\pi r^2\) to get \(h=\frac{4}{3}r\).

a) Both volumes are \(36\pi\,\text{cm}^3\). b) \(h=\frac{4}{3}r\)