Find volume bounded by 3 equations using integration

Question

The prompt is to find the volume of the solid bounded by the equations $x^2 + y^2 -2y = 0$, $z = x^2 + y^2$ and $z \ge 0$

Plotting the equations, I get something like this

We get a paraboloid and a cylinder.

How to know if I have to use double or triple integration to solve this problem? and if I should use spherical or double polar or cylindrical coordinate system? Also how to find the limits of integration in this particular problem as an example?

This is what the cross section looks like

It is clearly visible that $\theta$ goes from $0$ to $\pi$

Answer 1

$\textbf{Second method}$ (I presume that this is the method you want):

Show that the volume enclosed by the paraboloid $z=x^2+y^2$, the $xy$-plane $z=0$, and the cylinder $x^2+(y-1)^2=1$ is $\frac{3\pi}{2}$.

This time, use the change of coordinates: $$ \begin{align*} x &= r\cos \theta, \\ y &=r\sin\theta, \\ z &=z, \\ \end{align*} $$ where $$ \begin{align*} 0 \leq &z\leq r^2, \\ 0\leq &r \leq 2\sin\theta, \mbox{ and }\\ 0 \leq &\theta \leq \pi. \\ \end{align*} $$

Let $E$ be the solid defined in this prompt. Then the volume of the enclosed object is $$ \begin{align*} V&=\iiint_E 1 \: dV \\ &= \int_{0}^{\pi}\int_{0}^{2\sin\theta} \int_{0}^{r^2} r\: dzdrd\theta \\ &= \int_{0}^{\pi}\int_{0}^{2\sin\theta} r^3 \: dr d\theta \\ &= \int_{0}^{\pi} \frac{r^4}{4}\Bigg|_0^{2\sin\theta} \: d\theta \\ &= \int_{0}^{\pi} 4\sin^4\theta\: d\theta \\ &= \int_{0}^{\pi} 4\left( \frac{1-\cos(2\theta) }{2}\right)^2 \: d\theta \\ &= \int_{0}^{\pi} \left( 1-\cos(2\theta) \right)^2 \: d\theta \\ &= \int_{0}^{\pi} \left( 1-2\cos(2\theta)+\cos^2(2\theta) \right) \: d\theta \\ &= \int_{0}^{\pi} \left( 1-2\cos(2\theta)+\frac{1}{2}+\frac{1}{2}\cos(4\theta) \right) \: d\theta \\ &= \theta +\sin(2\theta)+\frac{1}{2}\theta +\frac{1}{8}\sin(4\theta) \Bigg|_0^{\pi} \\ &=\frac{3}{2}\pi + 0 + 0 \\ &=\frac{3}{2}\pi. \end{align*} $$

Answer 2

First, complete the square to see that $$ x^2+y^2-2y=x^2 +(y-1)^2-1=0. $$ So $x^2+(y-1)^2=1$. The cylindrical change-of-coordinate we will be using is: $$ x=r\cos \theta, \hspace{4mm} y=1+r\sin \theta,\hspace{4mm} z=z, \hspace{4mm} \mbox{ where } 0\leq r\leq 1,\hspace{4mm} 0\leq \theta\leq 2\pi. $$ Note that the absolute value of the Jacobian is: $$ \left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| = \left| \pmatrix{ x_r& x_\theta \\ y_r & y_\theta \\ } \right| = r. $$ Now, we calculate the volume of the solid: $$ \begin{align*} \iint_{D} \int_0^{x^2+y^2} dz\: dA &= \iint_{D} x^2+y^2 \: dA \\ &= \int_{0}^{2\pi}\int_{0}^{1} \left( (r\cos\theta)^2 + (1+r\sin \theta)^2 \right) r\: dr d\theta \\ &= \int_{0}^{2\pi}\int_{0}^{1} \left( r^2\cos^2\theta+ r^2\sin^2 \theta + 1+2 r\sin \theta \right) r\: dr d\theta \\ &= \int_{0}^{2\pi}\int_{0}^{1} \left( r^2 + 1+2 r\sin \theta \right) r\: dr d\theta \\ &= \int_{0}^{2\pi}\int_{0}^{1} \left( r^3 + r + 2 r^2 \sin \theta \right) dr d\theta \\ &= \int_{0}^{2\pi} \left( \frac{r^4}{4} + \frac{r^2}{2} + 2 \frac{r^3}{3} \sin \theta \right) \Bigg|_{0}^1 d\theta \\ &= \int_{0}^{2\pi} \left( \frac{1}{4} + \frac{1}{2} + \frac{2}{3} \sin \theta \right) d\theta \\ &= \frac{\pi}{2}+\pi -\frac{2}{3}\cos\theta\Bigg|_{0}^{2\pi} \\ &= \frac{3\pi}{2}. \end{align*} $$

Answer 3

$\textbf{Third method}$: find the volume of the cylinder, where $0\leq z\leq 4$, and then subtract from it the volume of the cylinder whose lower bound is the paraboloid and upper bound is the plane $z=4$.

So rather than working with the equations: $$ \color{blue}{z=0, \hspace{4mm} z=x^2+y^2, \hspace{4mm} \mbox{ and } \hspace{4mm} x^2+(y-1)^2=1,} $$ imagine picking up the solid and then moving it so that the axis of rotation for the cylinder $\underline{\textbf{is}}$ the $z$-axis. This solid is now bounded by the equations: $$ \color{green}{z=0, \hspace{4mm} z=x^2+(y+1)^2, \hspace{4mm} \mbox{ and } \hspace{4mm} x^2+y^2=1,} $$ which has been plotted below:

The volume of the cylinder is given by $$ \text{Volume}(\text{cylinder}) = \pi R^2 h = \pi 1^2(4) = 4\pi. $$ The volume of the cylinder $E'$ whose lower bound is the paraboloid and upper bound is the plane $z=4$ is $$ \begin{align*} \text{Volume}(E') &= \int_0^{2\pi} \int_{0}^{1}\int_{(r\cos\theta)^2+(r\sin\theta+1)^2}^{4}r\: dz dr d\theta \\ &= \int_0^{2\pi} \int_{0}^{1} 4r-\left( r^3\cos^2\theta + r(r\sin\theta+1)^2\right) \: dr d\theta \\ &= \int_0^{2\pi} \int_{0}^{1} \left( 3 r - r^3 (\cos^2\theta + \sin^2\theta )- 2 r^2 \sin\theta \right) \: dr d\theta \\ &= \int_0^{2\pi} \int_{0}^{1} \left( 3 r - r^3 - 2 r^2 \sin\theta \right) \: dr d\theta \\ &= \int_0^{2\pi} \left( \frac{3}{2} r^2 - \frac{1}{4}r^4 - \frac{2}{3} r^3 \sin\theta \right) \Bigg|_0^1 \: d\theta \\ &= \int_0^{2\pi} \left( \frac{5}{4} - \frac{2}{3} \sin\theta \right) \: d\theta \\ &= \frac{5}{4}\theta +\frac{2}{3} \cos\theta\Bigg|_0^{2\pi} \\ &= \frac{5}{2}\pi. \end{align*} $$ Thus $$ \text{Volume}(\text{solid}) = 4\pi - \frac{5}{2}\pi = \boxed{\frac{3}{2}\pi}. $$