# Triple integral - finding mass of solid

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

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