Disk and Washer Method

Applications of Integration

6. Volumes

• Understanding the concept of volume in calculus alongside finding areas.

6.2 Volumes

Volume of Solids

• Volume is defined precisely through calculus.

Definition of Volume (1 of 10)

• Right Cylinder: A solid formed by two congruent bases (plane regions) in parallel planes.

  • Let the bases be B1 and B2.

  • Volume formula: $V = A h$

  • Where A = area of the base, h = height of cylinder

Definition of Volume (2 of 10)

• Circular Cylinder: Base is a circle with radius r. Volume:
  $V = ext{Area of base} imes ext{height} = ext{Area of circle} imes h = ext{π r}^2 h$

• Rectangular Box: Base is a rectangle with length l and width w. Volume:
  $V = lwh$

Definition of Volume (3 of 10)

• For a solid S that isn’t a cylinder:

  • Cut S into pieces approximated by cylinders.

  • Sum the volumes of these cylinders to estimate the volume of S.

  • Use a limiting process as the number of pieces increases.

Definition of Volume (4 of 10)

• Cross-section: Intersect S with a plane to obtain a region corresponding to a cross-section.

  • Let $A(x)$ be the area of this cross-section at point x.

  • Varies as x ranges from a ≤ x ≤ b.

Definition of Volume (5 of 10)

• Divide S into n equal-width slabs (Δx).

  • Use planes $P<em>{x</em>1}, P<em>{x</em>2}, …$ to slice the solid.

  • Sample points $x<em>i^*$ in intervals $[x</em>{i-1}, x_i]$ approximate each slab (Si).

Definition of Volume (6 of 10)

• Each slab’s volume is approximated by $A(x_i^*) imes Δx$.

  • Sum of slabs gives an approximation of total volume.

Definition of Volume (7 of 10)

• As n -> ∞:

  • The approximation improves, yielding the volume as a limit of sums.

  • Recognizes this limit as a definite integral:
    $V = ext{lim}_{n o ext{∞}} ext{Sum of cylinder volumes}$

Definition of Volume (8 of 10)

• Recall that $A(x)$ is the area of a moving cross-section.

  • For cylinders, $A(x)$ is constant: $A(x) = A$ for all x, consistent with $V = Ah$.

Example 1: Volume of a Sphere

• To show the volume of a sphere of radius r:

  • Place the sphere at the origin. The cross-section at plane $P_x$ intersects in a circle.

  • Radius from the Pythagorean theorem: $r_{cross} = ext{sqrt}(r^2 - x^2)$

  • Therefore, the cross-sectional area is:
    $A(x) = ext{π (sqrt}(r^2 - x^2))^2 = ext{π (r}^2 - x^2)$

Example 1 – Solution

• Evaluating the volume from a = −r to b = r:
  $V = ext{∫}<em>{-r}^{r} A(x) dx = ext{∫}</em>{-r}^{r} ext{π (r}^2 - x^2) dx$.

Definition of Volume (9 of 10)

• Using $r = 1$, we approximate the volume of a sphere:

  • Volume: $V = rac{4}{3} ext{π r}^3 = rac{4}{3} ext{π}$.

Definition of Volume (10 of 10)

• Volume derived through Riemann sums with n = 5, 10, 20, observed with increasing refinement.

Volumes of Solids of Revolution

Volumes of Solids of Revolution (1 of 2)

• Revolving a region around a line produces a solid of revolution.

• Volume calculation:

  • For a disk cross-section, find the radius (in terms of x or y).
    $V = ext{π}A^2$ where A is the radius of the disk.

Volumes of Solids of Revolution (2 of 2)

• For a washer cross-section:

  • Calculate inner radius $r<em>{in}$ and outer radius $r</em>{out}$.

  • Area of the washer: $A = ext{π}(r<em>{out}^2 - r</em>{in}^2)$.

Example 6: Volume of a Solid of Revolution

• Rotate the region about the line x = −1:

  • Cross-section: Washer with inner and outer radii.

  • Calculate the area using the equation given in the context.

Example 6 – Solution

• The volume is given by:
  $V = ∫A(y) dy = ext{π} ∫[f(y) - g(y)] dy$.

Finding Volume Using Cross-Sectional Area

• Analysis of solids with non-revolutionary shapes having computable cross-sectional areas.

Example 7: Volume of a Solid with Circular Base

• Base radius = 1, cross-sections are equilateral triangles.

• The volume of the solid is determined from the cross-sectional area calculations.

Example 7 – Solution (1 of 3)

• Circle equation: $x^2 + y^2 = 1$, visualizing the solid and cross-section.

Example 7 – Solution (2 of 3)

• Cross-section area and triangle dimensions are computed based on geometry.

Example 7 – Solution (3 of 3)

• Final volume obtained through quantified integration.
  $V = ∫_{-1}^{1} A(x) dx$ where A(x) is calculated earlier.

Conclusion

• Integration serves as a powerful tool for volume determination across various solid shapes by taking into account their unique properties and cross-sections.