please do you mean it is a function and it's inverse the function and it's mirror by the line of angle 45 so the bounded areas for the same integral for the same period is equal as they will be in mirror mode also

* please do you mean it is a function and it's inverse the function and it's mirror by the line of angle 45 so the bounded areas for the same integral is equal as they will be in mirror mode also??*

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestConsider the area in the first quadrant bounded by \(x^7+y^4 = 1\) with \(x,y \ge 0\) and the coordinate axes.

Since the curve is given by \(y = \sqrt[4]{1-x^7}\), \(x \in [0,1]\), the area is \(\displaystyle\int_{0}^{1}\sqrt[4]{1-x^7}\,dx\).

Since the curve is given by \(x = \sqrt[7]{1-y^4}\), \(y \in [0,1]\), the area is \(\displaystyle\int_{0}^{1}\sqrt[7]{1-y^4}\,dy\).

But both methods must yield the same area. Therefore: \(\displaystyle\int_{0}^{1}\sqrt[4]{1-x^7}\,dx = \displaystyle\int_{0}^{1}\sqrt[7]{1-y^4}\,dy\).

Thus, \(\displaystyle \int_{0}^{1}\left[\sqrt[4]{1-x^7} - \sqrt[7]{1-x^4}\right]\,dx = \displaystyle\int_{0}^{1}\sqrt[4]{1-x^7}\,dx - \displaystyle\int_{0}^{1}\sqrt[7]{1-x^4}\,dx = 0\).

Log in to reply

I love this solution! Still, it left me wondering if there is more standard method (perhaps in multivariable calculus) that formalizes this?

Log in to reply

I'm not sure what you mean by formalize, but if you want to use multivariable calculus:

Let \(\Omega = \{(x,y) | 0 \le x \le 1 , 0 \le y \le \sqrt[4]{1-x^7}\} = \{(x,y) | 0 \le y \le 1 , 0 \le x \le \sqrt[7]{1-y^4}\}\).

Then, by Theorem III in this Wikipedia article, we have:

\(\displaystyle\iint\limits_{\Omega} \,dA = \int_{0}^{1} \int_{0}^{\sqrt[4]{1-x^7}} \,dy\,dx = \int_{0}^{1}\sqrt[4]{1-x^7} \,dx\)

\(\displaystyle\iint\limits_{\Omega} \,dA = \int_{0}^{1} \int_{0}^{\sqrt[7]{1-y^4}} \,dx\,dy = \int_{0}^{1}\sqrt[7]{1-y^4} \,dy\)

It's not hard to see that the conditions of that theorem are satisfied. The conclusion follows.

Log in to reply

what a great solution. Thanks!

Log in to reply

That's a nice one Jimmy, thanks for sharing!

Log in to reply

Nyc... this is an elegant solution..short and elegant.

Log in to reply

please do you mean it is a function and it's inverse the function and it's mirror by the line of angle 45 so the bounded areas for the same integral for the same period is equal as they will be in mirror mode alsoLog in to reply

* please do you mean it is a function and it's inverse the function and it's mirror by the line of angle 45 so the bounded areas for the same integral is equal as they will be in mirror mode also??*Log in to reply

Since the curve is given by y=1−x7−−−−−√4, x∈[0,1], the area is ∫101−x7−−−−−√4dx.

Since the curve is given by x=1−y4−−−−−√7, y∈[0,1], the area is ∫101−y4−−−−−√7dy.

But both methods must yield the same area. Therefore: ∫101−x7−−−−−√4dx=∫101−y4−−−−−√7dy.

Thus, ∫10[1−x7−−−−−√4−1−x4−−−−−√7]dx=∫101−x7−−−−−√4dx−∫101−x4−−−−−√7dx=0.

Log in to reply