Math Problem Statement

47–61 Find the volume of the described solid S. 47. A right circular cone with height h and base radius r using integration

Solution

To find the volume of a right circular cone using integration, we can slice the cone horizontally and sum up the volumes of thin disks with infinitesimal thickness.

Steps to Solve:

  1. Equation of the Cone:

    • A cone has a base radius rr and a height hh.
    • Let's position the cone so that its vertex is at the origin (0,0)(0,0) and its height extends along the yy-axis up to y=hy = h. The radius at the base (where y=hy = h) is rr.
    • At a height yy, the radius R(y)R(y) of the cross-section is proportional to yy. From similar triangles, the radius at any height yy is: R(y)=rhyR(y) = \frac{r}{h} y
  2. Volume of a Thin Disk:

    • Consider a thin slice of the cone at height yy with thickness dydy. The radius of this disk is R(y)=rhyR(y) = \frac{r}{h} y.
    • The volume dVdV of this thin disk is: dV=π[R(y)]2dy=π(rhy)2dydV = \pi [R(y)]^2 \, dy = \pi \left( \frac{r}{h} y \right)^2 \, dy dV=πr2h2y2dydV = \pi \frac{r^2}{h^2} y^2 \, dy
  3. Integrating to Find Total Volume:

    • To get the total volume of the cone, integrate the volume of the disks from y=0y = 0 to y=hy = h: V=0hπr2h2y2dyV = \int_0^h \pi \frac{r^2}{h^2} y^2 \, dy
  4. Performing the Integration:

    • The integral is: V=πr2h20hy2dyV = \pi \frac{r^2}{h^2} \int_0^h y^2 \, dy
      • The integral of y2y^2 is: 0hy2dy=y330h=h33\int_0^h y^2 \, dy = \frac{y^3}{3} \Big|_0^h = \frac{h^3}{3}
    • Substituting this back: V=πr2h2h33=πr2h3V = \pi \frac{r^2}{h^2} \cdot \frac{h^3}{3} = \pi \frac{r^2 h}{3}

Thus, the volume of the cone is: V=13πr2hV = \frac{1}{3} \pi r^2 h

This is the standard formula for the volume of a cone.


Let me know if you'd like further details or have any questions!

5 Related Questions:

  1. How would the method change if the cone was an oblique cone instead of a right cone?
  2. What would be the volume of a cone with a slant height instead of vertical height?
  3. How can you extend this integration approach to find the volume of a cone with an elliptical base?
  4. Can you derive the volume of a cylinder using a similar method of integration?
  5. How can you compute the surface area of this cone using integration?

Tip:

When solving problems involving solids of revolution, setting up the geometry carefully with correct limits and functions is crucial for successful integration.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Solid Geometry

Formulas

V = (1/3)πr²h (Volume of a cone)
R(y) = (r/h)y (Radius of cross-section at height y)

Theorems

Volume of Solids of Revolution

Suitable Grade Level

Grades 11-12 (High School Calculus)