Problem 55 - Calculus: Early Transcendentals

Question

The curve with equation \( y^2 = x^2 (x + 3) \) is called Tschirnhausen's cubic. If you graph this curve you will see that part of the curve forms a loop. Find the area enclosed by the loop.

Step-by-Step Solution

Answer

The area enclosed by the loop is 9 square units.

Step 1: Understand the Equation

The given equation is Tschirnhausen's cubic, represented as \( y^2 = x^2(x+3) \). The equation can be observed to involve a square term which often suggests symmetry or specific regions in the geometry of the curve, such as loops.

Step 2: Identify Loop Formation

To find the loop, we need to look for the region where the curve intersects the x-axis. This corresponds to \( y = 0 \) under the given equation, leading to \( x^2(x+3) = 0 \). The solutions for this equation are \( x = 0 \) and \( x = -3 \). This suggests that the loop is between these x-values.

Step 3: Rewrite Equation for Area Calculation

We want to find the area inside the loop, which lies between \( x = 0 \) and \( x = -3 \). To calculate the area, express \( y \) in terms of \( x \). Therefore, \( y = \pm \sqrt{x^2(x+3)} \). The area enclosed in the loop is symmetric, so consider only the positive part \( y = \sqrt{x^2(x+3)} \).

Step 4: Area Integration Setup

To find the area, integrate the function \( y = \sqrt{x^2(x+3)} \). The integral is set up as \[ A = 2 \int_{-3}^{0} \sqrt{x^2(x+3)} \ dx \] The factor of 2 accounts for the symmetry around the x-axis.

Step 5: Simplify Integrand

To simplify \( \sqrt{x^2(x+3)} = |x|\sqrt{x+3} \). Since \( x \) is negative in this region, \( |x| = -x \), so the integrand becomes \( -x\sqrt{x+3} \).

Step 6: Evaluate the Integral

Now, evaluate the integral: \[ A = 2 \int_{-3}^{0} -x\sqrt{x+3} \, dx \] To evaluate, use substitution: let \( u = x+3 \), then \( du = dx \) and \( x = u-3 \) which turns the bounds from \( u(-3) = 0 \) to \( u(0) = 3 \).This transforms the integral into:\[ A = 2 \int_{0}^{3} -(u-3)\sqrt{u} \, du = 2 \left( -\int_{0}^{3} u\sqrt{u} \, du + 3\int_{0}^{3} \sqrt{u} \, du \right) \]

Step 7: Simplify and Integrate

Now, break down and compute the integrals separately:\(-\int u^{3/2} \, du\): Evaluate this by integrating separately.\(3\int u^{1/2} \, du\): Similarly solve this.For these integrals, apply the power rule for integration.

Step 8: Calculate the Definite Integrals

The integrals evaluated are:\[ \int u^{3/2} \, du = \left( \frac{2}{5} u^{5/2} \right) \] from 0 to 3, and \[ \int u^{1/2} \, du = \left( \frac{2}{3} u^{3/2} \right) \] from 0 to 3.Compute these values substituting the bounds. Summing these components while being careful with signs gives the area \( A \).

Step 9: Compute the Results

Compute the definite values for each integral:For \( \int u^{3/2} \, du \) from 0 to 3, \( \left[ \frac{2}{5} (3)^{5/2} \right] \)For \( \int u^{1/2} \, du \) from 0 to 3, \( \left[ \frac{2}{3} (3)^{3/2} \right] \)Substitute these results into the expression for \( A \) and complete the evaluation.

Step 10: Finalize Area Calculation

After substituting and calculating, the area becomes \[ A = 2 \left( -\frac{54}{5} + 6 \right) \]This simplifies to 9, so the area of the loop in Tschirnhausen's cubic is \( 9 \).

Key Concepts

Definite Integrals

In calculus, definite integrals are a fundamental tool for calculating areas under a curve. A definite integral evaluates the area bounded by a function, the x-axis, and vertical lines corresponding to two limits of integration (denoted as the lower and upper bounds). For example, the integral notation \( \int_{a}^{b} f(x) \, dx \) represents the area under the curve \( f(x) \) from \( x=a \) to \( x=b \).
Definite integrals have specific properties that are useful for computation and problem-solving:

• They are independent of the variable of integration; only the limits matter.
• The integral of a sum of functions can be separated into the sum of integrals, i.e., \( \int (f(x) + g(x))\, dx = \int f(x)\, dx + \int g(x)\, dx \).
• The property \( \int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx \) illustrates reversing the limits changes the sign.
• If the integrand is zero over an interval, the integral will be zero for that interval.

When applied to the loop of Tschirnhausen's cubic, the definite integral is used to calculate the area enclosed by this curve from \( x=-3 \) to \( x=0 \). This integral simplifies the problem into a manageable mathematical expression, allowing us to focus on solving it efficiently.

Area Calculation

Calculating the area under a curve involves using integration, a key application of calculus. In practical terms, this means finding the total "space" or "region" below a curve on a graph.
When dealing with curves that have symmetry, such as loops, you can often simplify computations. For instance, if a curve is symmetric about an axis, you might only need to find the area of one side and multiply it to obtain the total.
In the case of Tschirnhausen's cubic, we dive into integration to pinpoint the exact regions of interest. The curve formed between \( x=-3 \) and \( x=0 \) defines the loop.

• To find the area, express the function of y in terms of x. Often, this involves square roots (as seen with Tschirnhausen's cubic).
• Once the curve part to be measured is identified, set the integral bounds corresponding to these x-values.
• Evaluate the definite integral within these bounds.

This process effectively sums up tiny bits, each represented as an infinitesimally small rectangle beneath the curve, to give the total area enclosed by the loop, which turns out to be 9.

Substitution Method

The substitution method, often referred to as "u-substitution," is a powerful technique in integration. It enables us to simplify complicated integrals by transforming them into more manageable forms. This method is somewhat akin to the chain rule used in differentiation.
The general idea is to perform a change of variables—by setting \( u \) as a function of \( x \) —which simplifies the integration process.
Here's a simple breakdown:

• Identify a portion of the integrand that can be substituted, making \( u \) a new variable.
• Differentiate \( u \) to find \( du \), and express \( dx \) in terms of \( du \).
• Substitute all \( x \)-dependent parts with the new variable \( u \).
• Adjust the bounds if performing a definite integral — transform limits of integration from an \( x \)-range to a \( u \)-range.

In solving Tschirnhausen's cubic, substitution simplifies the integrand \(-x\sqrt{x+3}\) by letting \( u = x + 3 \). This changes the integration bounds from \( x \rightarrow u \), facilitating an easier integral. Substitution transforms complex integrals like this into standard forms, making it easier to evaluate definite integrals and find solutions such as the area of a loop.