I'm trying to find the mass of the solid (constant density $\rho $) which is bounded by the surface $x=y^2$ and the planes $x=z, z=0, x=1.$. Now I'm having a hard time visualising/sketching it as you would for just a double integral.

I tried some stuff and came up with $y^2\leq x\leq 1, -1\leq y\leq 1, 0\leq z\leq x $ and so I got the mass as being given by the iterated integrals $$\int_{-1}^1dy\int_{y^2}^{1}dx\int_{0}^{x}\rho \ dz$$ but I'm not too sure about this.

[same as $$\int_{-1}^1\int_{y^2}^{1}\int_{0}^{x}\rho \ dzdxdy$$]

  • $\begingroup$ One of planes need correction $z=0$. $\endgroup$ – Nosrati Feb 22 '17 at 16:10
  • $\begingroup$ Thank you. I wrote it twice for some reason. $\endgroup$ – Anon Feb 22 '17 at 16:20
  • $\begingroup$ You keep writing your integrals in the wrong way. You cannot write $\int_{-1}^1 dy \int_{y^2}^1 dx$. Instead, write $\int_{-1}^1 \int_{y^2}^1 dx\;dy$ $\endgroup$ – Kuifje Feb 22 '17 at 16:22
  • $\begingroup$ It's just different notation that I see being used. Is what I put correct though? $\endgroup$ – Anon Feb 22 '17 at 16:26
  • $\begingroup$ Why $-1\leq y\leq1$ and isn't $0\leq y\leq1$.? The solid is symmetric.? $\endgroup$ – Nosrati Feb 22 '17 at 18:24

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