91Ó°ÊÓ

Calculating the area of a region under a curve involves using integration. When a region is bounded by a curve, the x-axis, and vertical lines, the integral allows us to find the total area in this space. In our exercise, we're looking at the function \( y = ax^2 \) from \( x = 0 \) to \( x = 2 \).
To set up the integral, we use the equation:

• \( \text{Area} = \int_{0}^{2} ax^2 \, dx \)

This equation estimates the area by summing up an infinite number of infinitely thin rectangles under the curve.
After setting up the integral, calculate it to find the area:

• \( \int_{0}^{2} ax^2 \, dx = a \left[ \frac{x^3}{3} \right]_0^2 \)
• This simplifies to \( a \left( \frac{8}{3} \right) \)

So, the area of our region is \( \frac{8a}{3} \), depending on the value of \( a \). This technique is foundational in calculus for finding areas bounded by curves.

Creating a solid of revolution involves rotating a 2D shape around an axis to form a 3D object, which is a fascinating way calculus connects shapes with volumes. In this context, we rotate the area under the curve \( y = ax^2 \) around the x-axis. This forms a symmetrical 3D shape.
The disk method is a common way to determine the volume of such solids. Imagine stacking tiny disks, or cylindrical slices, along the x-axis. Each disk's volume is calculated and summed to find the total volume.

• The formula using the disk method is: \( V = \pi \int_{0}^{2} (ax^2)^2 \, dx \)

This integral effectively adds up the volume of each cylindrical slice from \( x = 0 \) to \( x = 2 \). Understanding this method helps visualize how a flat area can create a volumetric solid when rotated around an axis.

In calculus, setting up an integral is crucial for solving problems related to areas and volumes. It's the first and often most important step. It translates a real-world question into a mathematical expression we can work with.
For area calculation, our setup was \( \int_{0}^{2} ax^2 \, dx \), which means we integrate the function \( ax^2 \) across \( x = 0 \) to \( x = 2 \). This setup ensures we're calculating the area between the curve, the x-axis, and the vertical boundaries.
For calculating volumes, the setup becomes a bit more involved. Using the disk method, the integral formula \( V = \pi \int_{0}^{2} (ax^2)^2 \, dx \) was used. Here, \((ax^2)^2\) represents the square of the function we're rotating, giving us the radius squared for each tiny disk.

• The factor of \( \pi \) accounts for the circular nature of these disks.

Understanding the role of these integral setups provides a powerful tool for tackling a range of problems in calculus. It transforms physical intuitive concepts into precise mathematical computations.