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Could you please teach me how to do this integral? My question is mainly about determining the limits of integration

The question is to find the volume of the solid enclosed by these three surfaces $$\iiint\limits_D dv \text{ over } D=\begin{cases}y^{2}+z^{2}=4ax\\x=3a\\y^{2}=ax\end{cases}$$

The following are my trails $$\int_{0}^{3a}dx\int_{\sqrt{ax}}^{2\sqrt{ax}}dy\int_{-\sqrt{4ax-y^{2}}}^{\sqrt{4ax-y^{2}}}dz $$ and using trigonometric substitution $$\int_0^{3a}dx\int_0^{2\sqrt{ax}}rdr\int_0^{2\pi}d\theta $$

thanks a lot for your answers and apologize for my bad expressions

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  • $\begingroup$ All the three equations define surfaces in $\mathbb R^3$, which are two-dimensional objects, hence do not make a three-dimensional $D$. You need either to replace the 'equals' signs with correct 'is bigger than' or 'is smaller than' to denote appropriate side of each surface, and then $D$ is an intersection of those three three-dimensional regions; or specify in text that $D$ is a region delimited by the three surfaces of given equations. $\endgroup$
    – CiaPan
    Commented Jun 11 at 11:28
  • $\begingroup$ Even if you meant: $y^2\color{red}{\le} ax$ I don't see how this translates to an integral $$\int_{\sqrt{ax}}^{2\sqrt{ax}}\,dy\,.$$ $\endgroup$
    – Kurt G.
    Commented Jun 11 at 11:31
  • $\begingroup$ @KurtG. I tried to project the figure onto xoy plane. So z =0 and from the first and the third equals I get the integral $\endgroup$
    – imaretard
    Commented Jun 11 at 11:36

1 Answer 1

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Let's take a closer look at $D$. I assume $a>0$. The paraboloid $y^2+z^2=4ax$ intersects with the parabolic cylinder $y^2=ax$ in the planes $z^2=3y^2$, which follows from eliminating $x$. (intersection highlighted in blue)

enter image description here

Consider the cross section of $D$ at $x=3a$. Viewing the $(y,z)$ plane from above we would see a symmetric region like this:

enter image description here

Suppose we wanted to find the area of the region bounded by the orange and blue curves in polar coordinates, $(y,z)\to(r\cos\theta,r\sin\theta)$. This is given by the double integral

$$\int_0^{2\pi} \int_0^{r(\theta)} r \, dr \, d\theta = 2 \int_\alpha^{\pi+\alpha} \int_0^{r(\theta)} r \, dr \, d\theta$$

where $\alpha$ is any of the angles made by the planes $y^2=z^2$ with the positive $y$-axis.

However, $r(\theta)$ is not uniform as $\theta$ varies. If we pick $\alpha$ to be the angle made by the top-right vertex, for instance, the complete picture is captured by

$$r(\theta) = \begin{cases}\sqrt3\,a \lvert\sec\theta\rvert & \theta\in[0,\alpha]\cup[\pi-\alpha,\pi+\alpha]\cup[2\pi-\alpha,2\pi] \\ 2\sqrt3\,a & \theta\in[\alpha,\pi-\alpha]\cup[\pi+\alpha,2\pi-\alpha] \end{cases}$$

So we need to split up the integral at an appropriate value of $\theta$. By symmetry, we can reduce the area integral to $4$ times the integral over the region in the first quadrant, so that the area would be

$$4 \left\{\int_0^\alpha \int_0^{\sqrt3\,a\sec\theta} + \int_\alpha^\tfrac\pi2 \int_0^{2\sqrt3\,a}\right\} r \, dr \, d\theta$$


Now how might you extend this analysis to the volume integral?

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