Question

In Exercises 35-38, use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by the curves about the y-axis. $$y=\sqrt{x}, \quad y=0, \quad x=4$$

Short answer

The volume of the solid generated by revolving the given region around the y-axis is \(8\pi\).

Step-by-step solution

Identify the Parameters

Firstly, identify the limits of integration along the y-axis and the radius and height of every cylindrical shell. In this case, the limits are given as y = 0 and y = \( \sqrt{4} = 2\). The radius of each tiny cylindrical shell is the y value because we are revolving around the y-axis, which is y. The height of each cylindrical shell is given by the x value for a given y, because the region is bounded by the curve \(y=\sqrt{x}\), so \(x = y^2\).

Set Up the Integral

Now set up the integral for the volume using the formula for cylindrical shell method, which is \(V = 2 \pi \int_a^b yf(y) \, dy\). Here, \(a = 0\), \(b = 2\), and \(f(y) = y^2\), because \(x = y^2\). Therefore, the volume integral becomes \(V = 2 \pi \int_0^2 y . y^2 \, dy\).

Calculate the Integral

Evaluate the integral by calculating the antiderivative of the integrand and subtracting the result with lower limit from the result with upper limit. The antiderivative of \(y . y^2\) is \(\frac{y^4}{4}\). Therefore, \(V = 2 \pi [(\frac{2^4}{4}) - (\frac{0^4}{4})]\).

Final Computation

Perform arithmetic to find the final value of V. This results in \(V = 2 \pi [4 - 0] = 8\pi\).

Key concepts

Volume of Solids of Revolution

When studying calculus, one comes across a fascinating application which is finding the volume of solids of revolution. This involves creating a 3D shape by revolving a 2D area around a given axis. To visualize this, imagine a piece of paper that represents an area in two dimensions. Now, if you spin this paper around a pen acting as the axis, it would form a solid shape.

Mathematically, the volume can be computed by different techniques, with the cylindrical shell method being one such technique. In our exercise, the region under the curve of the function \(y = \sqrt{x}\), above the x-axis \(y = 0\), and between \(x = 0\) and \(x = 4\), is being revolved about the y-axis, giving rise to a solid whose volume can be calculated by integrating cylindrical shells.

Integrating Functions

Integrating functions is a cornerstone concept in calculus, representing the process of finding the integral of a function, which can be thought of as the 'reverse' of differentiation. Integrals can be definite, with specific limits, or indefinite, representing a general form. In the context of finding volumes, definite integrals are typically used.

To compute the volume of a solid of revolution, we need to integrate over the specific range that the solid occupies. In our example, we have used integration to sum up an infinite number of infinitesimally thin cylindrical shells to arrive at the total volume of the solid.

Cylinder Volume Calculus

The volume of a cylinder in calculus isn't just about using the formula \(V = \pi r^2h\); it involves understanding how to use calculus to find the volume of cylindrical shapes, especially when they change in size throughout their height, which is why the integration process comes into play.

Each cylindrical shell's volume can be thought of as the product of its circumference, its height, and an infinitesimally small thickness. For a cylindrical shell with radius 'r' and height 'h', the volume is given by \(dV = 2\pi rh \, dy\), where \(dy\) is the thickness of the shell. Then, integration is used to sum these volumes over a range to get the total volume.

Method of Shells

The method of shells is a technique for calculating the volume of solids of revolution, particularly when revolving around an axis other than the axis of the function. This method involves slicing the solid perpendicular to the axis of revolution, creating cylindrical shells, and adding up their volumes.

In our exercise, the cylindrical shells have a radius equal to their distance from the y-axis (y), and a height equal to the value of the function at that specific y (which, for the curve \(y = \sqrt{x}\), is \(x = y^2\)). The shell method formula \(V = 2\pi \int_a^b yf(y) \, dy\) allows us to set up an integral that gives us the total volume when evaluated, where 'a' and 'b' represent the limits of integration along the axis of revolution, and \(f(y)\) is the function that represents the height of the shell.