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I want to compute the volume of the solid that lies between the $xz$-plane, the $yz$-plane , the $xy$-plane, the planes with equations $x=1$ and $y=1$ and the surface with equation $z=x^2+y^4$.

Is it as follows?

It must hold that $0 \leq x \leq 1$, $0 \leq y \leq 1$.

So the volume of the solid described above is $\int_0^1 \int_0^1 (x^2+y^4) dxdy$. Is this right?

Also how can we compute the volume of the solid that lies between the plane with equation $z=16$ and the plane with equation $z=x^2+y^2$ ?

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  • $\begingroup$ " .. and the plane with equation ..." should be "and the surface with equation ..", isn't it ? $\endgroup$ – G Cab Oct 4 '17 at 12:01
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The volume you've given for the first question is correct.

To solve the second, notice that the curves overlap on the circle $z = 16, x^2+y^2=16.$

This means to calculate the area between the curves, simply integrate the following:

$$\int^4_{-4}\int^{\sqrt{16-x^2}}_{-\sqrt{16-x^2}}\left(16-x^2-y^2\right)dydx$$

$16-x^2-y^2$ represents the area between the plane and the paraboloid given. Imagine the original shapes and the shape between them, then compare that to $16-x^2-y^2.$

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  • $\begingroup$ Why do we look for the volume between the plane with equation $z=16$ and the plane with equation $z=x^2+y^2$ of the function $16-x^2-y^2$ ? @AustinWeaver I haven't understood it. $\endgroup$ – Evinda Oct 4 '17 at 8:27
  • $\begingroup$ in this case, $x^2+y^2 \le 16$ . I subtract them to get the difference, or length between the plane and curve at each $xy$ point. $\endgroup$ – Austin Weaver Oct 4 '17 at 11:12

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