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Chapter 3: Problem 4

Find the area of loop \(\mathrm{y}^{2}=\mathrm{x}(\mathrm{x}-1)^{2}\).

Short Answer

Expert verified
Answer: \(\frac{4}{3}\)

Step by step solution

01

Identify the type of graph

The given equation is \(y^2 = x(x-1)^2\). Since this equation isn't a simple algebraic expression, we know that it is a special type of graph. Let's solve for y to get a better understanding of the graph: \(y = \pm\sqrt{x(x-1)^2}\).
02

Determine the range of x values for the loop

Since \(y^2 = x(x-1)^2\), we know that the root of the graph will be \(x = 0\). We also see that another root is when \(x = 1\). These will be our limits of integration.
03

Set up the integral to compute the area

We want to compute the area between the curve and the x-axis. Remember that we have two different graphs since we have \(y = \pm\sqrt{x(x-1)^2}\). So, the area of the loop can be calculated as: $$ Area = 2\int_{0}^{1}\sqrt{x(x-1)^2}dx $$ The factor 2 is added to account for both positive and negative half of the graph.
04

Evaluate the integral

Now we evaluate the integral: $$ Area = 2\int_{0}^{1} \sqrt{x(x-1)^2} dx = 2\int_{0}^{1} x\cdot(x-1) dx $$ To evaluate this integral, we can use the power rule: $$ Area = 2\left[\frac{1}{2}x^2\cdot\frac{1}{3}(x-1)^{3}\right]_0^1 = 2\left(\frac{1}{2}\cdot\frac{1}{3}(1-1)^{3}-\frac{1}{2}\cdot\frac{1}{3}(0-1)^{3}\right) = \frac{4}{3} $$
05

End Result

The area of the loop found in the equation \(y^2 = x(x-1)^2\) is \(\frac{4}{3}\).

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Most popular questions from this chapter

Find the area included between curves \(\mathrm{y}=\frac{4-\mathrm{x}^{2}}{4+\mathrm{x}^{2}}\) and \(5 y=3|x|-6 .\)

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Compute the area of the curvilinear trapezoid bounded by the \(x\)-axis and the curve \(y=x-x^{2} \sqrt{x}\).
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