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I had a test today and want to know whether I set the limits right for this volume computation. The solid is the one defined by $$z= 2-2x^2-y^2, x + y + z = 3, 2x^2 + y^2 = 2$$, and the limits I set are (in adapted cylindrical coordinates in which

$$x=r\cos t , y = \sqrt{2}r\sin t, z = z$$ and the Jacobian $J = \sqrt{2}r$: $$ 0\le t \le 2\pi , 0\le r \le 1 , 2-2r^2 \le z \le 3-r\cos t - \sqrt{2}r\sin t$$

I got the volume $ V = 2\sqrt{2}\pi$

Can anyone verify this please?

Thank you so much!!

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We can visualize the three surfaces as follows: enter image description here

Looking at the image, we want the volume that is above the upside-down paraboloid, inside the elliptic cylinder, and below the plane.

What you are doing is not really converting to cylindrical coordinates, because the coordinate transformation is not: $$ x = r \cos \theta \qquad y = r \sin \theta \qquad z = z $$ rather we define a new coordinate transformation: $$ x = r \cos \theta \qquad y = \sqrt{2} r \sin \theta \qquad z = z$$ in which case, our volume of integration now corresponds to: $$ 0 \leq r \leq 1 \qquad 0 \leq \theta \leq 2\pi \qquad 2 - 2r^2 \leq z \leq 3 - r \cos \theta - \sqrt{2} r \sin \theta $$

and the Jacobian is: \begin{align*} \frac{\partial(x,y,z)}{\partial(r,\theta,z)} &= \frac{\partial(x,y)}{\partial(r,\theta)} \\ &= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} \\ &= \begin{vmatrix} \cos \theta & -r \sin \theta \\ \sqrt{2} \sin \theta & \sqrt{2} r \cos \theta \end{vmatrix} \\ &= \sqrt{2}\begin{vmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} \\ &= \sqrt{2}r \end{align*} So $$ dV = dx \, dy \, dz = \sqrt{2}r \, dr \, d\theta \, dz $$

So our volume integral becomes: \begin{align*} V &= \iiint \, dV \\ &= \int_0^{2\pi} \int_0^1 \int_{2-2r^2}^{3 - r \cos \theta - \sqrt{2}r \sin \theta} \sqrt{2}r \, dz \, dr \, d\theta \\ &= \sqrt{2}\int_0^{2\pi} \int_0^1 r\left(1 + 2r^2 - r\cos \theta - \sqrt{2} r\sin \theta \right) \, dr \, d\theta \\ &= \sqrt{2}\int_0^{2\pi} \int_0^1 (r+2r^3) \, dr \, d\theta - \sqrt{2}\int_0^{2\pi} \int_0^1 r^2\left(\cos \theta + \sqrt{2} \sin \theta \right) \, dr \, d\theta \\ &= 2\sqrt{2} \pi \int_0^1 (r+2r^3) \, dr - \sqrt{2}\int_0^{2\pi} \frac{1}{3}\left(\cos \theta + \sqrt{2} \sin \theta \right) \, d\theta \\ &= 2\sqrt{2} \pi - \frac{\sqrt{2}}{3}\int_0^{2\pi} \left(\cos \theta + \sqrt{2} \sin \theta \right) \, d\theta \\ &= 2\sqrt{2} \pi - 0 \\ &= 2\sqrt{2} \pi \end{align*}

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  • $\begingroup$ Thank you! In the upper limit for z, wouldn't it be 3 - ... instead of 2 -...? $\endgroup$ – Bee May 19 '17 at 4:40
  • $\begingroup$ Oh yea. Sorry about that. Corrected. $\endgroup$ – Manuel Guillen May 19 '17 at 8:11
  • $\begingroup$ Thanks! Also, could you please tell me what program you used for the 3D plot? It looks great! $\endgroup$ – Bee May 19 '17 at 11:10
  • $\begingroup$ Wolfram Mathematica $\endgroup$ – Manuel Guillen May 19 '17 at 17:07

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