Find volume where the region is bounded by $z=1-4(x^2+y^2)$ and the xy plane

I used cylindrical coordinate system to setup the integral $1-4(x^2+y^2)\geq z \geq 0$ $\Rightarrow$ $1-4r^2\geq z \geq 0$ and projection on xy plane is $x^2 + y^2 = 1/4$ or $r=1/2$

$\int_{0}^{2 \pi} \int_{0}^{1/2} \int_{0}^{1-4r^2} rdzdrd{\theta}$

on solving this I am getting $\frac {\pi}{8}$ as answer but the book lists $\frac{7\pi}{2}$ as answer $\cdots$ what am I doing wrong?

  • $\begingroup$ I double-checked by working as a solid of revolution and also got $\frac{\pi}{8}$. $\endgroup$ – John Wayland Bales Oct 5 '17 at 18:34
  • $\begingroup$ This is set up perfectly. $\endgroup$ – Ted Shifrin Oct 5 '17 at 18:41
  • $\begingroup$ Thanks guys. Probably misprinted answer at the back. $\endgroup$ – So Lo Oct 5 '17 at 19:48
  • $\begingroup$ @JohnWaylandBales thank you so much $\endgroup$ – So Lo Oct 5 '17 at 19:49

I think your answer is correct. Probably the book is wrong.


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