Introduction to Volume & Disk Method: Videos & Practice Problems

In this section, we explore how to find the volume of a three-dimensional solid using integrals, building on the concept of calculating area under a curve. To visualize this, we can think of a solid with a base defined by the function \( f(x) = 4 - x^2 \) over the interval from 0 to 2. The solid has square cross-sections perpendicular to the x-axis.

To find the volume, we begin by slicing the solid into thin, rectangular prisms. Each slice has a square cross-section, where the side length of the square is determined by the function \( f(x) \). The area \( A(x) \) of the square cross-section can be expressed as:

\[ A(x) = f(x) \times f(x) = (4 - x^2)^2 \]

Next, to find the volume of a single slice, we multiply the area of the cross-section by the width of the slice, which is \( dx \). Thus, the volume \( dV \) of one slice is given by:

\[ dV = A(x) \cdot dx = (4 - x^2)^2 \cdot dx \]

To find the total volume of the solid, we integrate this expression over the interval from 0 to 2:

\[ V = \int_{0}^{2} (4 - x^2)^2 \, dx \]

This integral will yield the total volume of the solid. It is essential to visualize both the solid and its cross-section to avoid errors in setup and calculation. As you work through the integral, remember that the process of slicing the solid and summing the volumes of the slices is analogous to finding the area under a curve, but extended into three dimensions.

In summary, the key steps involve identifying the base function, determining the area of the cross-section, and integrating over the specified interval to find the total volume. This method can be applied to various shapes and cross-sections, enhancing your understanding of volume calculations in calculus.

In this discussion, we explore the evaluation of a volume integral for a three-dimensional shape formed by extending a curve upwards, where the cross sections are squares. The area of a square is calculated as the side length squared, leading us to integrate the function \( f(x)^2 \) to find the volume.

To set up the integral, we first square the function, resulting in the expression:

\[ \int_{0}^{2} (16 - 8x^2 + x^4) \, dx \]

Next, we find the antiderivative of each term within the integral:

\[ 16x - \frac{8}{3}x^3 + \frac{1}{5}x^5 \bigg|_{0}^{2} \]

Applying the Fundamental Theorem of Calculus, we evaluate this from 0 to 2. Plugging in the upper limit:

\[ 16(2) - \frac{8}{3}(2^3) + \frac{1}{5}(2^5) \]

Calculating each term gives:

• \( 16 \times 2 = 32 \)
• \( 2^3 = 8 \) so \( \frac{8}{3} \times 8 = \frac{64}{3} \)
• \( 2^5 = 32 \) so \( \frac{1}{5} \times 32 = \frac{32}{5} \)

Thus, the expression simplifies to:

\[ 32 - \frac{64}{3} + \frac{32}{5} \]

To combine these fractions, we find a common denominator of 15:

• \( 32 = \frac{480}{15} \)
• \( \frac{64}{3} = \frac{320}{15} \)
• \( \frac{32}{5} = \frac{96}{15} \)

Now, we can combine the fractions:

\[ \frac{480}{15} - \frac{320}{15} + \frac{96}{15} = \frac{256}{15} \]

This fraction represents the volume of the three-dimensional shape with square cross sections. If desired, this can also be expressed as a decimal, approximately \( 17.07 \). This volume is significant as it quantifies the space occupied by the shape defined by the given function.

If there are any questions regarding the setup or evaluation of this integral, further clarification is encouraged.

To find the volume of a three-dimensional solid formed by rotating a function around an axis, we utilize the concept of solids of revolution. In this context, the cross-sections of the solid are circular, which simplifies the process of determining the area function we need to integrate. The area of a circle is given by the formula \( A = \pi r^2 \), where \( r \) is the radius of the circle.

Consider the function \( f(x) = 4 - x^2 \) defined on the interval from \(-2\) to \(2\). When this function is revolved around the x-axis, it creates a solid of revolution. The radius \( r \) of the circular cross-section at any point \( x \) is determined by the distance from the curve to the axis of rotation. In this case, since we are revolving around the x-axis (where \( y = 0 \)), the radius function is simply the value of the function itself, \( r(x) = f(x) = 4 - x^2 \).

To set up the volume integral, we integrate the area of the circular cross-section from the lower limit to the upper limit of the interval. The volume \( V \) can be expressed as:

\( V = \int_{-2}^{2} \pi (r(x))^2 \, dx = \int_{-2}^{2} \pi (4 - x^2)^2 \, dx \)

This integral represents the volume of the solid formed by revolving the function around the x-axis. To find the numerical value of the volume, one would evaluate this integral. The process involves expanding the integrand, integrating term by term, and then applying the limits of integration.

Understanding the relationship between the function, its radius, and the resulting volume is crucial in mastering the concept of solids of revolution and the disk method. This foundational knowledge will be beneficial as you continue to explore more complex shapes and integration techniques in calculus.

To find the volume of a solid formed by revolving a function around the x-axis, we utilize the method of disks. The volume can be calculated using the integral of the area of circular cross-sections. For a function defined from -2 to 2, the radius of each circular cross-section is determined by the function itself, specifically r(x) = 4 - x². The area of each circular cross-section is given by the formula A = πr², leading to the integral:

V = ∫_-2² π(4 - x²)² dx.

To evaluate this integral, we first expand the integrand:

(4 - x²)² = 16 - 8x² + x⁴.

This allows us to rewrite the volume integral as:

V = π ∫_-2² (16 - 8x² + x⁴) dx.

Next, we can factor out the constant π and compute the integral:

V = π [16x - (8/3)x³ + (1/5)x⁵]_-2².

Applying the Fundamental Theorem of Calculus, we evaluate the antiderivative at the bounds:

V = π [16(2) - (8/3)(2³) + (1/5)(2⁵)] - π [16(-2) - (8/3)(-2³) + (1/5)(-2⁵)].

Calculating the upper bound:

16(2) = 32, (8/3)(2³) = 64/3, (1/5)(2⁵) = 32/5.

For the lower bound:

16(-2) = -32, (8/3)(-2³) = 64/3, (1/5)(-2⁵) = -32/5.

Combining these results, we simplify the expression:

V = π [32 - 64/3 + 32/5 + 32 + 64/3 - 32/5].

Combining like terms yields:

V = π [64 - (128/3) + (64/5)].

To combine the fractions, we find a common denominator, which is 15:

64 = 960/15, 128/3 = 640/15, 64/5 = 192/15.

Thus, we have:

V = π [(960 - 640 + 192)/15] = π [512/15].

Finally, the volume of the solid of revolution is:

V = (512π)/15.

This result represents the volume of the solid formed by revolving the given function around the x-axis.