Question

Find the volume of the solid generated when the curve \(y=3 x^{2}\) for \(0 \leq x \leq 1\) is rotated around the \(y\) axis.

Short answer

Answer: The volume of the solid is \(\dfrac{3}{2}\pi\) cubic units.

Step-by-step solution

Write the integral for the volume

The integral for the volume using the method of cylindrical shells is given by: \(V = 2\pi\int_0^1{x \cdot 3x^2 dx}\) Now we need to solve the integral to find the volume.

Evaluate the integral

First, simplify the integrand by multiplying \(x\) with \(3x^2\): \(V = 2\pi\int_0^1{3x^3 dx}\) Next, find the antiderivative of \(3x^3\): \(\int{3x^3 dx} = \dfrac{3}{4}x^4 + C\) Now, we will evaluate the definite integral using the limits \(0 \leq x \leq 1\): \(V = 2\pi\left[\dfrac{3}{4}(1)^4 - \dfrac{3}{4}(0)^4\right]\)

Calculate the volume

Now we just need to plug in the values and find the volume: \(V = 2\pi\left[\dfrac{3}{4} - 0\right] = \dfrac{3}{2}\pi\) Therefore, the volume of the solid generated when the curve \(y=3x^2\) is rotated around the \(y\)-axis is \(\dfrac{3}{2}\pi\) cubic units.

Key concepts

Cylindrical Shells Method

Understanding the cylindrical shells method is crucial when calculating the volume of a solid of revolution, especially when the axis of rotation is parallel to the y-axis. This method imagines the solid as being composed of numerous thin 'shells'. Each cylindrical shell is hollow, like a pipe segment, with the wall's thickness being an infinitely small value, typically represented by \( dx \) or \( dy \) depending on the axis of rotation.

To calculate the volume of each shell, we use the formula for the volume of a cylinder (\[ V_{shell} = 2\text{π}\text{radius}\text{height}\text{thickness} \]). In the problem at hand, the shells' radii are the distances from the y-axis (and can be thought of as the x-coordinates), the heights are given by the function (in this case, \( y = 3x^2 \) ), and the thickness of each shell is an increment along the x-axis (represented by \( dx \) ).

The integration of these volumes across the specified interval \( [0, 1] \) will result in the total volume of the solid. It's a beautiful application of integration to approach a complex three-dimensional problem through one-dimensional slices.

Definite Integral

A definite integral is a concept from calculus that is used to find the signed area under a curve. When it comes to the volume of solids of revolution, the definite integral accumulates the infinitesimal contributions of each cylindrical shell's volume.

In our exercise, \( \text{2π}\text{∫}_0^1{x \text{3x}^2 dx} \) represents the accumulation of all of the small cylindrical shells from \( x = 0 \) to \( x = 1 \) — the area we are interested in.

Evaluating the Definite Integral

Within the context of this problem, evaluating the definite integral requires finding the antiderivative of the function within the integrand and then applying the limits of integration. We often represent this evaluation as:
\[ V = F(b) - F(a) \]
where \( F(x) \) is the antiderivative and \( a \) and \( b \) are the limits of the definite integral.

Volume Calculation

Finally, the volume calculation brings together the cylindrical shells method and the definite integral. After evaluating the definite integral, we are left with the expression \( V = 2\text{π}\frac{3}{4} \), which needs to be simplified to provide the volume of the solid of revolution. Plugging in the numbers, we find that the volume is \( \frac{3}{2}\text{π} \) cubic units.

It's important to realize that this calculation gives us a concrete volume for an abstractly defined shape, which we conceptualize without having to physically construct the object. The beauty of mathematics is how it enables these theoretical constructs and provides exact answers for otherwise intangible concepts.

Understanding each step of this volume calculation, from the concept of using cylindrical shells to the execution of the definite integral, allows students to engage with complex geometry and calculus problems in a manageable and logical way.