I'm studying Double and Triple Integrals and stucked understanding how to determinate which reduction formula is ok for my case. Also got problems in finding $\rho$ and $\theta$ while switching in polar coordinates. Tried to solve this :
- Calculate the volume |A|, with A = {$x$ $\in$ $\mathfrak R^3$ : $z$ $\ge$ $x^2+$ $y^2$, $z \le 1$, $x\ge z$}
By setting $y = 0$, I drew in $z$ and $x$ axes functions to know which region to look and :
Region is under green line $z=x$ and above $z=x^2$ in blue
Then :
$$\iiint_A dxdydz = \iint_Ddxdy\int_{x^2+y^2}^xdz = \iint_D(x-x^2-y^2)dxdy$$
To determinate $D$ i'm setting $x = x^2 + y^2$ who, by manipulation, is $(x-\frac{1}{2})^2 +$ $y^2$ = $\frac{1}{4}$ So a circle with center = $(\frac{1}{2},0)$ and radius = $\frac{1}{2}$
Now by setting polar coordinates I end up with $0 \le\rho\le cos\theta$ and $0\le\theta\le\frac{\pi}{4}$.
My final answer to this is $\frac{8+3\pi}{384}$. Result is different and it's $\frac{3\pi}{128}$.
I need a comparison just to know if I'm doing this thing in a right way.
Thank you for all your assistance.
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As you can see, the region '$x > 1$' has empty intersection with '$x^{2} + y^{2} < z < 1$'. So, your integration is solely over a region with '$x < 1$' and '$x^{2} + y^{2} < z < x$'.
\begin{align} \iiint_{A}\dd x\,\dd y\,\dd z & = \iiint_{\large\mathbb{R}^{3}}\bracks{x^{2} + y^{2} \leq z \leq \min\braces{1,x}} \dd x\,\dd y\,\dd z \\[1cm] & = \iiint_{\large\mathbb{R}^{3}}\bracks{x \leq 1} \bracks{x^{2} + y^{2} \leq z \leq x}\dd x\,\dd y\,\dd z \\[5mm] & + \iiint_{\large\mathbb{R}^{3}}\bracks{x \geq 1} \bracks{x^{2} + y^{2} \leq z \leq 1}\dd x\,\dd y\,\dd z \\[1cm] & = \int_{-\infty}^{\infty}\int_{-\infty}^{1}\bracks{x^{2} + y^{2} \leq x} \int_{x^{2} + y^{2}}^{x}\dd z\,\dd x\,\dd y \\ & + \overbrace{\int_{-\infty}^{\infty}\int_{1}^{\infty} \bracks{x^{2} + y^{2} \leq 1} \int_{x^{2} + y^{2}}^{1}\dd z\,\dd x\,\dd y}^{\ds{=\ 0}} \\[1cm] & = \int_{-\infty}^{\infty}\int_{-\infty}^{1/2} \bracks{x^{2} + y^{2} \leq {1 \over 4}} \int_{x^{2} + x + 1/4 + y^{2}}^{x + 1/2}\dd z\,\dd x\,\dd y = \int_{0}^{2\pi}\int_{0}^{1/2}\pars{{1 \over 4} - r^{2}}r\,\dd r\,\dd\phi \\[5mm] & = \bbx{\ds{\pi \over 32}} \end{align}