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I've tried calculating the volume limited by the surfaces $y=z^2$, $z=x^2$, $x=y^2$, $2y=z^2$, $2z=x^2$, $2x=y^2$. I didn't know how to begin so instead I tried solving for $y=z^2$, $z=x^2$, $x=y^2$.

So I checked where they intersect and they can only have values from $0$ to $1$. I projected over the $xy$-plane and got: $$\iint_D\left(\int_0^{x^2}\mathrm dz\right)\,\mathrm dx\mathrm dy=\int_0^1\left(\int_0^{\sqrt x}x^2\,\mathrm dy\right)\,\mathrm dx=\frac25.$$

I don't think that I got this right, not really confident in what I did. If what I did is correct, how should I proceed to resolve the original problem now?

Ѕᴀᴀᴅ
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iggykimi
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    Do you want to find the volume bounded by the first set of six surfaces or the second set of three surfaces? – Ѕᴀᴀᴅ Apr 26 '20 at 11:43
  • The first one @Saad I believe that if I see how to do that one, I'll be able to do the second. – iggykimi Apr 26 '20 at 18:06
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    Hi! if you want to do the second one, first think about the projections of these three surfaces on $xy,yz,xz$-planes and try to imagine the region(volume) bounded by these three surfaces in space and think the surfaces which are bounding $z$ from the below and from the above, then imagine the projection of the surfaces (intersection)on $xy$-plane, you will get the limits of $x,y$ this way. This will give the volume:$$V=\int_{x=0}^{1}\int_{y=x}^{\sqrt{x}}\int_{z=x^2}^{\sqrt{y}}dzdydx.$$ – Learning Apr 27 '20 at 19:08

1 Answers1

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$\def\R{\mathbb{R}}\def\e{\mathrm{e}}\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}$Assume that the bounded volume is$$ A = \left\{ (x, y, z) \in \R^3 \,\middle|\, \dfrac{y^2}{2} \leqslant x \leqslant y^2,\ \dfrac{z^2}{2} \leqslant y \leqslant z^2,\ \dfrac{x^2}{2} \leqslant z \leqslant x^2\right\} $$ and make the substitution $(x, y, z) = (\e^u, \e^v, \e^w) =: f(u, v, w)$. Since $\dfrac{∂(x, y, z)}{∂(u, v, w)} = \exp(u + v + w)$ and\begin{align*} B := f^{-1}(A) &= \{(u, v, w) \in \R^3 \mid 2v - \ln 2 \leqslant u \leqslant 2v,\\ &\peq 2w - \ln 2 \leqslant v \leqslant 2w,\ 2u - \ln 2 \leqslant w \leqslant 2u\}, \end{align*} then$$ \iiint\limits_A \d x\d y\d z = \iiint\limits_B \left| \frac{∂(x, y, z)}{∂(u, v, w)} \right| \,\d u\d v\d w = \iiint\limits_B \exp(u + v + w) \,\d u\d v\d w. $$ Now make the substitution $(r, s, t) = (2u - w, 2v - u, 2w - v) =: g(u, v, w)$. Because $r + s + t = u + v + w$, $$ \frac{∂(u, v, w)}{∂(r, s, t)} = \left( \frac{∂(r, s, t)}{∂(u, v, w)} \right)^{-1} = \begin{vmatrix} 2 & 0 & -1\\ -1 & 2 & 0\\ 0 & -1 & 2 \end{vmatrix}^{-1} = \frac{1}{7} $$ and$$ C := g(B) = \{(r, s, t) \in \R^3 \mid 0 \leqslant r, s, t \leqslant \ln 2\}, $$ so\begin{gather*} \iiint\limits_B \exp(u + v + w) \,\d u\d v\d w = \iiint\limits_C \exp(r + s + t) \left| \frac{∂(u, v, w)}{∂(r, s, t)} \right| \,\d r\d s\d t\\ = \frac{1}{7} \iiint\limits_{0 \leqslant r, s, t \leqslant \ln 2} \e^r \e^s \e^t \,\d r\d s\d t = \frac{1}{7} \left( \int_0^{\ln 2} \e^t \,\d t \right)^3 = \frac{1}{7}, \end{gather*} i.e. the volume is $\dfrac{1}{7}$.

Ѕᴀᴀᴅ
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