Problem 4 - Thomas Calculus

Question

In Exercises \(1 - 8 :\) a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve to see what it looks like. If you can, graph the surface too. c. Use your utility's integral evaluator to find the surface's area numerically. $$x = \sin y , \quad 0 \leq y \leq \pi ; \quad y -axis$$

Step-by-Step Solution

Answer

The area is obtained by evaluating the integral using \(x = \sin y\) revolved about the \(y\)-axis.

Step 1: Understand Revolved Surface

We are given a curve described by the equation \(x = \sin y\) and need to find the area of the surface generated by revolving this curve about the \(y\)-axis over the interval \(0 \leq y \leq \pi\). The key is recognizing this forms a surface of revolution.

Step 2: Establish Integral Setup

When a curve is revolved about the \(y\)-axis, the area of the surface generated is given by the integral \(2\pi \int_a^b x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \ dy\). Here, \(x = \sin y\), so first compute \(\frac{dx}{dy}\).

Step 3: Compute Derivative for Integration

Calculate \(\frac{dx}{dy}\) for the curve \(x = \sin y\). Since \(dx/dy = \cos y\), substitute this into the surface area integral formula.

Step 4: Set Up the Specific Integral

Substituting \(x = \sin y\) and \(\frac{dx}{dy} = \cos y\) into the integral formula, we have:\[ 2\pi \int_0^\pi \sin y \sqrt{1 + (\cos y)^2} \ dy. \] Simplify \(1 + \cos^2 y\) for further processing.

Step 5: Simplify and Evaluate Integral Numerically

Solve the integral \(2\pi \int_0^\pi \sin y \sqrt{1 + \cos^2 y} \ dy\) using a calculator or numerical integration tool like a graphing calculator or software capable of numerical computations.

Key Concepts

Integral Calculus

Integral calculus helps us find quantities like areas, volumes, and other things by accumulating tiny bits of information.
It's especially useful when dealing with curves and complicated shapes. Here, it's the tool that allows us to compute the surface area of a revolved curve.
The integral we need comes from a formula specifically for surfaces formed by rotating a curve around an axis.When a curve, like the one given in this exercise, is spun about the y-axis, it forms a special kind of surface called a surface of revolution. The formula for finding the area of such a surface is:

• \( A = 2\pi \int_a^b x \, \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy \)

This formula is central to the solution.First, note that the curve is described by \( x = \sin y \) and this is revolved around the y-axis. To apply the formula, we calculate the derivative \( \frac{dx}{dy} = \cos y \).
Then, we substitute the function and its derivative into the integral to find the surface area.

Surface Area

Calculating surface area for shapes created by revolving a curve is slightly different than simple areas.
Because the shape involves a 3D surface, the formula for a surface of revolution integrates circular slices of the surface.
After finding the integral expression, it's about taking the curve \( x = \sin y \) from the interval \( 0 \leq y \leq \pi \) and revolving it.The formula mentioned earlier, breaks down to:

• The term \( 2\pi \) accounts for the circular nature of each slice.
• The integral \( \int_0^\pi \sin y \sqrt{1 + \cos^2 y} \, dy \) sums up all the tiny surface area elements.

All these parts combine to give the total surface area. To find the answer, using numerical methods such as calculators or software can evaluate the integral to a specific value.

Revolved Curve

A revolved curve in mathematics is a way to create a 3D shape by spinning a flat curve around a line—usually one of the axes.
The curve given by \( x = \sin y \) in this exercise is spun around the y-axis, which forms a surface called a surface of revolution.
How exactly does this work? Imagine if you drew a wave on a piece of paper and then twirled that piece of paper around.
You'd end up with a 3D shape. What revolution of the curve does is similar, but more precise.For the curve \( x = \sin y \) from \( y = 0 \) to \( y = \pi \), spinning it around the y-axis creates a kind of bell shape.
This type of problem not only involves understanding the geometry of the situation but also requires calculation skills.
By revolving curves, we can find the surface area or volume of many different shapes and uses tools from calculus to handle computation effectively.