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Find the limits of integration for $\iiint_Ef(x,y,z)dzdydx$ where $E$ is bounded by the paraboloid $x=4y^2 + 4z^2$ an the plane $x=4$

I understand that it forms a paraboloid along the x axis. If we draw it on the xy plane where $z = 0$, we find that $x = 4y^2$, which is a parabola along the x axis limited by the line $x=4$. $$ \int_{-1}^1\int_{4y^2}^4\int f(x,y,z)\:dz\:dy\:dx $$

If we view it along the z axis, the paraboloid forms a circle where $x=4$, such that $4 = 4y^2 + 4z^2$. Solving for $z$ we get: $z = \pm\sqrt{1-y^2}$

So the limits of integration I came up with were:

$$ \int_{-1}^1\int_{4y^2}^4\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} f(x,y,z)\:dz\:dy\:dx $$

Where did I go wrong?

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1 Answer 1

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To start with, the possible values of $x$ clearly range from $0$ to $4$, not from $-1$ to $1$. The limits on your middle integral make no sense: that’s the integral with respect to $y$, so it definitely should not have $y$ in the limits of integration. In fact it should have only constants and $x$, since at that point the $z$ has already been integrated out. Finally, the limits on your innermost integral, with respect to $z$, can’t be right, since the range of possible values for $z$ depends on both $x$ and $y$, not just on $y$.

The intersection of the region with the plane $x=a$ is bounded by $4y^2+4z^2=a$: projected onto the $yz$-plane, it’s the region inside the circle $y^2+z^2=\frac{a}4$ of radius $\sqrt{a}/2$ centred at the origin. Thus, when $x=a$, $y^2$ ranges from a minimum of $0$ to a maximum of $\frac{a}4$, and $y$ ranges from $-\frac12\sqrt{a}$ to $\frac12\sqrt{a}$. Finally, if $x=a$ and $y=b$, then the bounds on $z^2$ are $0$ and $\frac{a}4-b^2$, so the bounds on $z$ are $\pm\sqrt{\frac{a}4-b^2}$. Putting the pieces together gives us the integral

$$\int_0^4\int_{-\sqrt{x}/2}^{\sqrt{x}/2}\int_{-\sqrt{\frac{x}4-y^2}}^{\sqrt{\frac{x}4-y^2}}f(x,y,z)~dzdydx\;.$$

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