Find the volume of the solid bounded by

Question

Find the volume of the solid bounded by $x^2+y^2–2y = 0, z = x^2+y^2 , z = 0 $. I have to calculate volume using triple integrals but I struggle with finding intervals. I calculated $x_1=\sqrt{2y-y^2}$ and $x_2 = -\sqrt{2y-y^2}$. I think $z$ will have interval from $0$ to $2y$, but I don't know what to do next.

Answer 1

Since $x^2+y^2-2y=0\iff x^2+(y-1)^2=1$, you know that $\lvert y-1\rvert\leqslant1$ and that therefore $0\leqslant y\leqslant2$. So, if you try to solve the problem in cylindrical coordinates, $0\leqslant\theta\leqslant\pi$. On the other hand$$z=x^2+y^2\iff z=r^2$$and$$x^2+y^2=2y\iff r=0\vee r=2\sin\theta.$$So, compute$$\int_0^\pi\int_0^{2\sin\theta}\int_0^{r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$

Answer 2

The first equation, $x^2+(y-1)^2=1$, is the two dimensional surface of a cylinder in $\Bbb R^3$, it cuts the $xOy$ plane in the circle with the same equation, centered in $(0,1)$ and radius one. So the axis $Oz$ is one generating line of the cylinder. Draw this object now in 3D, although we will need only the two dimensions.

Then the part delimited by the cylinder, that may lead to a compact body is $$ x^2+(y-1)^2\le 1^2\ . $$ In the $z$ direction we go from $z=0$ to $z=x^2+y^2$. So we have to integrate: $$ J = \iint_{x^2+(y-1)^2\le 1^2}(x^2+y^2-0)\; dx\; dy\ . $$ Let us substitute $x=r\cos t$, $y=1+r\sin t$. Then $$ \begin{aligned} J &= \int_0^1\int_0^{2\pi}(r^2\cos^2t +(1+r\sin t)^2)\; r\; dr\; dt \\ &= \int_0^1\int_0^{2\pi}(r^2 + 1 + 2r\sin t)\; r\; dr\; dt \\ &= \int_0^1\int_0^{2\pi}(r^2 + 1)\; r\; dr\; dt \\ &= 2\pi\int_0^1(r^2 + 1)\; r\; dr \\ &= 2\pi\left[\ \frac 14r^4+\frac 12r^2\ \right]_0^1 \\ &=2\pi\cdot\frac 34\ . \end{aligned} $$