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Given the title (i.e., find the volume of the solid $E$ bounded by the surfaces $z = 0$, $z = y$ and $y = 1 − x^2$), I've characterized the bounded region $E$ of interest as $$E=\{(x,y,z) \, \vert -1 \leq x \leq 1, \, 0 \leq y \leq 1-x^2, \, 0 \leq z \leq y \}$$

Which, provided my characterization above is correct (I'm not entirely sure about that), is the small red region bounded by the two green regions in the picture below:

enter image description here

I'm stuck trying to evaluate:

$$ \iiint_E y \,\,\mathrm{d}V$$

because it inevitably turns into

$$\int_{-1}^1 (1-x^2)^3 \, \mathrm{d}x$$

once you work out the other two innermost integrals, after which you're left with above which is a very tedious integral to solve. I've been told there is an easier way, though I don't know how!

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1 Answer 1

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$$\int_{-1}^{1}\int_{0}^{1-x^2}\int_{0}^{y} dzdydx= \int_{-1}^{1}\int_{0}^{1-x^2}ydydx= \frac{1}{2}\int_{-1}^{1}(1-x^2)^2dx = \\= \frac{1}{2}\int_{-1}^{1}(1-2x^2+x^4)dx=\frac{8}{15}$$

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  • $\begingroup$ Why is your integrand 1? I thought the integrand was y, seeing as we're integrating below z=y? $\endgroup$
    – Ius Klesar
    Commented Jul 3, 2020 at 0:29
  • $\begingroup$ When you calculate volume you take $f = 1$ and limits in 3d. In our case limits against $z$ are $0 \leqslant z \leqslant y$. $\endgroup$
    – zkutch
    Commented Jul 3, 2020 at 0:34

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