Question

\(x=2 y^{2}-y^{3}, x=0\), and \(y=0\) rotated around the \(x\) -axis using cylindrical shells

Short answer

The volume is \( \frac{16\pi}{5} \).

Step-by-step solution

Set up the equations

We are given the equations \( x = 2y^2 - y^3 \) and \( x = 0 \) with bounds \( y=0 \) rotated around the \( x \)-axis. The equation \( x = 2y^2 - y^3 \) defines a function of \( y \).

Determine the bounds of integration

Solve for \( x = 0 \) in the equation \( 2y^2 - y^3 = 0 \). Factoring gives \( y^2(2-y) = 0 \), leading to roots \( y = 0 \) and \( y = 2 \). Thus, the limits of integration are \( y = 0 \) to \( y = 2 \).

Set up the integral for volume using cylindrical shells

The formula for volume using cylindrical shells is \( V = \int_{a}^{b} 2\pi y (f(y)) dy \). Here, \( f(y) = 2y^2 - y^3 \), so the integral becomes \( V = \int_0^2 2\pi y (2y^2 - y^3) dy \).

Simplify the integrand

Expand the integrand: \( 2\pi y (2y^2 - y^3) = 2\pi (2y^3 - y^4) = 4\pi y^3 - 2\pi y^4 \).

Integrate

Integrate each term separately: \[ \\int 4\pi y^3 \, dy = 4\pi \cdot \frac{y^4}{4} = \pi y^4, \\int -2\pi y^4 \, dy = -2\pi \cdot \frac{y^5}{5} = -\frac{2\pi y^5}{5} \\]Thus, the integral becomes \( V = \left[ \pi y^4 - \frac{2\pi y^5}{5} \right]_0^2 \).

Evaluate the definite integral

Evaluate at the limits:At \( y = 2 \): \( V = \pi(2)^4 - \frac{2\pi(2)^5}{5} = 16\pi - \frac{64\pi}{5} \).At \( y = 0 \):\( V = 0 \).So, \( V = 16\pi - \frac{64\pi}{5} = \frac{80\pi}{5} - \frac{64\pi}{5} = \frac{16\pi}{5} \).

Conclusion

The volume of the solid obtained by rotating the curve \( x = 2y^2 - y^3 \) from \( y=0 \) to \( y=2 \) around the \( x \)-axis is \( \frac{16\pi}{5} \).

Key concepts

integration volume

In calculus, finding volumes of solids of revolution is a key skill. One common method is using integration to calculate volume. When a shape is generated by rotating a curve around an axis, we can use integration to determine the volume. The volume of these solids can be challenging to visualize, but using the method of cylindrical shells simplifies things. When dealing with the given curve equations, such as \( x = 2y^2 - y^3 \), integration helps us find the volume by taking slices of the solid—these slices are essentially the shells.

As we perform the integration, we consider how each shell contributes to the overall volume of the solid. The method involves setting up an integral that corresponds to the volume of these shells. In our exercise, each shell has a height \( f(y) = 2y^2 - y^3 \) and a radius \( y \), which rotates around the \( x \)-axis.

• Integration sets up the framework for adding up infinitely small contributions to the total volume.
• Integral bounds define the range of \( y \), in this case from \( y = 0 \) to \( y = 2 \).
• You compute the exact volume by evaluating this integral within these bounds.

Integration fundamentally transforms a geometrical problem into a calculative one, making complex shapes more manageable.

rotating curves

Rotating curves to form solids is a fascinating concept that bridges geometry and calculus. Imagine taking a two-dimensional graph of a function and revolving it around an axis to create a three-dimensional solid. In our example, we used the equation \( x = 2y^2 - y^3 \), which we rotated around the \( x \)-axis. This type of rotation leads to the generation of a solid shape, whose volume we aim to find.

The mechanics of rotating a curve to generate a solid is straightforward:

• Identify the curve to rotate, which in this case is given as a function of \( y \).
• Determine the axis of rotation, which affects the integration method—here, it's the \( x \)-axis.
• The rotation creates cylindrical sections (or shells), which are then summed up to get the volume.

The idea here is converting a planar area under the curve to a solid shape by spinning it around the axis, which is an excellent way to understand the relationship between dimensions. Through this method, complex volumes can be broken down into simpler integrating problems.

definite integrals

Definite integrals play a crucial role in precisely calculating volumes of solids formed by rotating curves. The integral provides a way to accumulate the contributions of each infinitesimal cylindrical shell to the total volume. In this specific exercise, the definite integral is set up as \( V = \int_0^2 2\pi y (2y^2 - y^3) \, dy \), where each term is carefully defined.

A definite integral consists of:

• Integration limits: These define where the integration begins and ends, in our case from \( y = 0 \) to \( y = 2 \).
• Integrand: The expression \( 2\pi y (2y^2 - y^3) \) represents the volume of each infinitesimally thin cylindrical shell formed as the curve is rotated.

Evaluating the definite integral solidifies the calculated volume. Each term in the expression \( 4\pi y^3 - 2\pi y^4 \) contributes incrementally as you integrate from the lower to the upper limit.

The step-by-step evaluation, substituting end limits, ensures an accurate final volume, in this case, resulting in \( \frac{16\pi}{5} \). This process exemplifies how definite integrals serve as a powerful tool in solving real-world volume problems.