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i'm facing the following problem: find the area of the set $M=\{(x,y): |x|^{\frac{2}{3}}+|y|^{\frac{2}{3}}\le 1\}$ using integration

i thought about only integrate where x,y both $\ge0$ and multiply by 4 using the function: $(t, (1-x^{\frac{2}{3}})^{\frac{3}{2}})$ thus i need to solve

$4\int_0^1 \! \sqrt{(1-x^{\frac{2}{3}})^3} \, \mathrm{d}x. $

what would be the right direction? i have been trying substitution using sin() function as used for $1-x^2$ examples, but it doesnt help much

i'm sure there is a simple trick to get started here

thanks for any hint on this

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3 Answers 3

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The substitution $x=(\sin t)^3$ surely leads to the solution.

(The title and the body of your post are asking different questions. This answer works for both.)

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I think that the first thing I should do is to get rid of the exponent $2/3$. So, starting with $x=y^3$, $dx=3y^2dy$ $$\sqrt{1-x^{2/3}}dx=3y^2\sqrt{1-y^{2}}dy$$ Now, the substitution $y=\sin(u)$ becomes quite obvious.

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Same as Claude Leibovici's

I think that the first thing I should do is to get rid of the exponent $2/3$. So, starting with $x=y^3$, $dx=3y^2dy$ $$\sqrt{1-x^{2/3}}dx=3y^2\sqrt{1-y^{2}}dy$$

except instead of substitution $y=\sin(\theta)$, you can consider evaluating the area of half of a semi circle.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Mar 19, 2018 at 7:19
  • $\begingroup$ @Taroccoesbrocco edited. thanks! $\endgroup$
    – BCLC
    Mar 19, 2018 at 7:32

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