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Chapter 3: Problem 17

Prove that the area between the curve \(\left(\frac{\mathrm{x}}{\mathrm{a}}\right)^{2 / 3}+\frac{\mathrm{y}}{\mathrm{b}}=1\) and the segment \((-\mathrm{a}, \mathrm{a})\) of the axis of \(x\) is \(\frac{4}{5} a b\).

Short Answer

Expert verified
Question: Prove that the area between the curve \((\frac{x}{a})^{2/3} + \frac{y}{b} = 1\) and the segment \((-a, a)\) of the x-axis is \(\frac{4}{5}ab\). Answer: We proved that the area between the given curve and the x-axis segment from -a to a is indeed \(\frac{4}{5}ab\) by following these steps: 1. We found the function of y in terms of x: \(y = b \left(1 - (\frac{x}{a})^{2/3}\right)\) 2. We integrated this function with respect to x from -a to a. 3. We evaluated the integral to obtain: \(ab[2-\frac{6}{5}]\) 4. We compared the result with the given area \(\frac{4}{5}ab\) and found that they are equal, thus proving the statement.

Step by step solution

01

Finding the function of y in terms of x

First, isolate y in the equation. \((\frac{x}{a})^{2/3} + \frac{y}{b} = 1 \Rightarrow y = b \left(1 - (\frac{x}{a})^{2/3}\right)\)
02

Integrating y with respect to x

Integrate the function of y with respect to x from -a to a to find the area. \(\int_{-a}^{a} b\left(1 - (\frac{x}{a})^{2/3}\right) dx = b\int_{-a}^{a}\left(1 - (\frac{x}{a})^{2/3}\right) dx\) Use substitution method: Let \(u = \frac{x}{a} \Rightarrow x = au \Rightarrow dx = adu\) Hence, our integral becomes: \(b\int_{-1}^{1}\left(1 - u^{2/3}\right) a du = ab\int_{-1}^{1}\left(1 - u^{2/3}\right) du\)
03

Evaluating the integral

Evaluate the integral: \(ab\int_{-1}^{1}\left(1 - u^{2/3}\right) du = ab\left[ u - \frac{3}{5}u^{\frac{5}{3}}\right]_{-1}^{1}\) Now, substitute the values -1 and 1 as the limits of the integration: \(ab\left[\left(1-\frac{3}{5}\cdot1^{\frac{5}{3}}\right) - \left(-1-\frac{3}{5}\cdot(-1)^{\frac{5}{3}}\right)\right] = ab[2-\frac{6}{5}]\)
04

Comparing the result with the given area

Now compare the result obtained above with the given area: \(Area = ab[2-\frac{6}{5}] = \frac{4}{5}ab\) Thus, we proved that the area between the curve \((\frac{x}{a})^{2/3} + \frac{y}{b} = 1\) and the segment \((-a, a)\) of the x-axis is \(\frac{4}{5}ab\).

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