To prove the formula for the volume of a cone algebraically, we need to derive the volume formula, $V = \frac{1}{3} \pi r^2 h$, where $r$ is the radius of the cone's base and $h$ is the height of the cone.

Derivation using calculus (integration):

Step 1: Understanding the geometry of the cone

Consider a cone with:

• A base of radius $r$,
• A height $h$, and
• A vertical axis along the $z$-axis.

The volume of the cone can be viewed as the sum of the volumes of thin circular slices taken from the base to the tip along the height of the cone.

Step 2: Setting up the integral

Imagine slicing the cone horizontally at a height $z$ from the tip. At this height, the radius of the cone's cross-section is proportional to $z$ since the cone tapers linearly from its base to the tip.

The radius of a cross-section at height $z$ can be expressed as:

$r(z) = r \cdot \frac{h - z}{h}$

The area of a cross-section at height $z$ is a circle with radius $r(z)$, so the area is:

$A(z) = \pi [r(z)]^2 = \pi \left( r \cdot \frac{h - z}{h} \right)^2$

Step 3: Volume of an infinitesimally thin slice

An infinitesimal slice of the cone at height $z$ has a small thickness $dz$, and its volume is the product of the area of the slice and the thickness:

$dV = A(z) \, dz = \pi \left( r \cdot \frac{h - z}{h} \right)^2 dz$

Step 4: Integrating to find the total volume

To find the total volume of the cone, we integrate this expression from $z = 0$ to $z = h$ (from the tip of the cone to the base):

$V = \int_0^h \pi \left( r \cdot \frac{h - z}{h} \right)^2 dz$

Simplifying the integrand:

$V = \pi r^2 \int_0^h \left( \frac{h - z}{h} \right)^2 dz$

$V = \frac{\pi r^2}{h^2} \int_0^h (h - z)^2 dz$

Step 5: Evaluating the integral

To evaluate $\int_0^h (h - z)^2 dz$, we expand the square:

$(h - z)^2 = h^2 - 2hz + z^2$

Now, integrate term by term:

$\int_0^h (h^2 - 2hz + z^2) dz = \int_0^h h^2 \, dz - 2h \int_0^h z \, dz + \int_0^h z^2 \, dz$

Each of these integrals is standard:

$\int_0^h h^2 dz = h^2 z \Big|_0^h = h^3$ $\int_0^h z \, dz = \frac{z^2}{2} \Big|_0^h = \frac{h^2}{2}$ $\int_0^h z^2 dz = \frac{z^3}{3} \Big|_0^h = \frac{h^3}{3}$

So, the total integral is:

$h^3 - 2h \cdot \frac{h^2}{2} + \frac{h^3}{3} = h^3 - h^3 + \frac{h^3}{3} = \frac{h^3}{3}$

Step 6: Final result

Now substitute this result back into the expression for the volume:

$V = \frac{\pi r^2}{h^2} \cdot \frac{h^3}{3} = \frac{1}{3} \pi r^2 h$

Thus, the volume of the cone is:

$V = \frac{1}{3} \pi r^2 h$

Conclusion

We have proven algebraically that the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.

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Do you have any questions about the steps? Would you like more details?

Here are five related questions:

1. How does the volume of a cone compare to the volume of a cylinder with the same base and height?
2. Can we derive the volume of a cone using geometry instead of calculus?
3. What is the volume of a cone with a slant height instead of the perpendicular height?
4. How does the volume formula for a truncated cone differ from that of a regular cone?
5. What is the relationship between the surface area and volume of a cone?

Tip: For 3D objects like cones, integrals are a powerful tool to calculate volumes by summing infinitesimal slices.