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Chapter 3: Problem 6

Find the area bounded by the curve \(x^{4}+y^{4}=x^{2}+y^{2}\)

Short Answer

Expert verified
Question: Determine the area enclosed by the curve \(x^{4}+y^{4}=x^{2}+y^{2}\). Solution: The area enclosed by the given curve can be found by calculating the integral \(A_{total} = 4\int_0^1 \sqrt{\frac{-x^{4}+x^{2}}{1}} dx\). This integral is complex and might require a calculator or software to compute the final result.

Step by step solution

01

Rewrite the equation in terms of y as a function of x

First, rewrite the given equation \(x^{4}+y^{4}=x^{2}+y^{2}\), as \(y^{4}-y^{2}+x^{4}-x^{2}=0\). Now, let's find y as a function of x. We see that it is a quadratic equation in terms of \(y^{2}\): \(y^{2}=\frac{-x^{4}+x^{2}}{1}\) To find the area under the curve, we will need to find the positive value for y: \(y=\sqrt{\frac{-x^{4}+x^{2}}{1}}\)
02

Calculate the x-boundaries (Where the curve intersects the x-axis)

Since we know that the curve will intersect the x-axis when y=0, we can find the values of x for which \(y=\sqrt{\frac{-x^{4}+x^{2}}{1}}=0\). This occurs when \(x^{4}=x^{2}\), or when \(x^{2}(x^{2}-1)=0\). This means that the curve intersects the x-axis at x=0 and x=1. Since the curve is symmetric, we are interested in integrating from 0 to 1.
03

Integrate y(x), from 0 to 1 to find the area in one quadrant

Now, we will integrate y with respect to x to find the area enclosed by the curve in the first quadrant: \(A_{1} = \int_0^1 \sqrt{\frac{-x^{4}+x^{2}}{1}} dx\)
04

Multiply A_1 by 4 to get the total area enclosed by the curve

Since the curve is symmetric about both the x-axis and the y-axis, we can multiply our result for A_1 by 4 to obtain the total area enclosed by the curve: \(A_{total} = 4A_{1} = 4\int_0^1 \sqrt{\frac{-x^{4}+x^{2}}{1}} dx\) It is important to note that the integral in its current form is quite complex to solve analytically, and the integration might need to be computed numerically using software or a calculator.

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Most popular questions from this chapter

For what value of a does the straight line \(\mathrm{y}=\mathrm{a}\) bisects the area of the figure bounded by the lines \(y=0, y=2+x-x^{2} ?\)

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