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I have to find the volume cut off from the paraboloid $4z=x^2+y^2$ by the plane $z=4$

I understand that the paraboloid will have its vertex at the orgin and its axis as the $z-axis$. now, if it is cut off by the plane $z=4$, i get the following limits for the triple intgral

$z$ from $0$ to $4$

$x$ from $-\sqrt{16-y^2}$ to $\sqrt{16-y^2}$

$y$ from $-16$ to $16$

However, this is double the correct answer. Where am i going wrong ?

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We can interpret this volume as $$\int_0^4 \text{area of disk at height } z\ dz.$$ More formally, we convert to a cylindrical coordinate system, $$\int_0^4 \underbrace{\int_0^{2\pi} \int_0^{\sqrt{4z}} r\ dr\ d\theta}_{\text{area of disk}}\ dz.$$

The area of a disk is $\pi r^2$ where $r$ is the radius, in this case $r=\sqrt{4z}$, so the area is $4\pi z$, and the integral becomes $$\int_0^4 4\pi z\ dz=\left[2\pi z\right]_0^4=32\pi.$$


Alternatively, we can find the volume of solid of revolution via $$2\pi \int_0^{4} x|f(x)-g(x)|\ dx$$ where $f(x)=4$ and $g(x)=\tfrac{1}{4}x^2$. This gives the volume \begin{align*} 2\pi \int_0^{4} 4x-\tfrac{1}{4}x^3\ dx &= 2\pi \left[2x^2-\tfrac{1}{16} x^4\right]_0^{4} \\ &= 2\pi(2 \times 4^2-\tfrac{4^4}{16}) \\ &= 32\pi. \end{align*}

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  • $\begingroup$ That was very understandable. Thanks !! can you please also point out what is wrong with my interpretation ? $\endgroup$ – Aman Mittal Sep 22 '13 at 15:56
  • $\begingroup$ I'm not entirely sure how it's meant to work. But the terminals on the $x$ and $y$ variables seem incorrect, as they should vary with $z$. $\endgroup$ – Rebecca J. Stones Sep 22 '13 at 16:07
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$x$ actually goes from $-\sqrt{4 z-y^2}$ to $\sqrt{4 z-y^2}$. Thus the volume is

$$V=\int_0^4 dz \, \int_{-\sqrt{4z}}^{\sqrt{4 z}} dy \, \int_{-\sqrt{4 z-y^2}}^{\sqrt{4 z-y^2}} dx$$

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  • $\begingroup$ why cant we take the projection of first equation on $z=4$ and then integrate ? $\endgroup$ – Aman Mittal Sep 22 '13 at 11:41
  • $\begingroup$ @AmanMittal: because that is a cylinder, not a paraboloid. $\endgroup$ – Ron Gordon Sep 22 '13 at 11:42
  • $\begingroup$ then what is $x^2+y^2=2x$ in three dimensions ? $\endgroup$ – Aman Mittal Sep 22 '13 at 11:49
  • $\begingroup$ @AmanMittal: a cylinder. $\endgroup$ – Ron Gordon Sep 22 '13 at 11:52

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