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PostPosted: Sun Sep 24, 2017 6:37 pm 
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Let $a,b,c,d$ positive real numbers such that $a<b$, $d^2=\dfrac{b-a}{2c}$ and the closed disk \[D=\big\{(x,y)\in{\mathbb{R}}^2\;\big|\;(x-x_0)^2+(y-y_0)^2\leqslant d^2\big\}\,.\] Find the volume and the surfase area of the solid \[V=\Big\{(x,y,z)\in{\mathbb{R}}^3\;\big|\;(x,y)\in{D}\,,\; a+c(x-x_0)^2+c(y-y_0)^2\leqslant{z}\leqslant b-c(x-x_0)^2-c(y-y_0)^2\Big\}\,.\]

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PostPosted: Sun Nov 26, 2017 8:02 am 
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$D=\big\{(x,y)\in{\mathbb{R}}^2\;\big|\;(x-x_0)^2+(y-y_0)^2\leqslant d^2\big\}$ is a closed disk with center $(x_0,y_0)$ and radius $d=\sqrt{\frac{b-a}{2c}}$. The paraboloids $z=a+c(x-x_0)^2+c(y-y_0)^2$, $z=b-c(x-x_0)^2-c(y-y_0)^2$, by which the solid \[V=\Big\{(x,y,z)\in{\mathbb{R}}^3\;\big|\;(x,y)\in{D}\,,\; a+c(x-x_0)^2+c(y-y_0)^2\leqslant{z}\leqslant b-c(x-x_0)^2-c(y-y_0)^2\Big\}\,,\] is defined, have as intersection the circle $\big\{(x,y,z)\in{\mathbb{R}}^3\;\big|\;(x-x_0)^2+(y-y_0)^2= d^2\,,\; z=\frac{a+b}{2}\big\}$. So, the solid $V$ has the form
Attachment:
aplogIV_5.png
aplogIV_5.png [ 78.21 KiB | Viewed 89 times ] aplogIV_5.png [ 78.21 KiB | Viewed 89 times ]

$V$ is normal set in ${\mathbb{R}}^3$, with respect to $xy$-plane. So, the volume of $V$ is
\begin{align*}
V(V)&= \mathop{\iint}\limits_{D}b-c(x-x_0)^2-c(y-y_0)^2-\big(a+c(x-x_0)^2+c(y-y_0)^2\big)\,d(x,y)\\
&= \mathop{\iint}\limits_{D}b-a-2c(x-x_0)^2-2c(y-y_0)^2\,d(x,y)\\
&= \frac{1}{2c}\int_{-d}^{d}\int_{x_0-\sqrt{d^2-(x-x_0)^2}}^{x_0+\sqrt{d^2-(x-x_0)^2}}\cancelto{d^2}{\frac{b-a}{2c}}-(x-x_0)^2-(y-y_0)^2\,dy\,dx\\
&\stackrel{(*)}{=} \frac{1}{2c}\int_{0}^{d}\int_{0}^{2\pi}(d^2-r^2)\,|r|\,d\varphi\,dr\\
&= \frac{\pi}{c}\int_{0}^{d}d^2r-r^3\,dr \\
&= \frac{\pi}{c}\,\frac{d^4}{4}\\
&= \frac{d^4\pi}{4c}\,.
\end{align*} $(*)$ Change of coordinates \begin{align*}
\left({
\begin{array}{c}
x(r,\varphi)\\
y(r,\varphi)
\end{array}
}\right)&=\left({
\begin{array}{c}
x_0+r\cos\varphi\\
y_0+r\sin\varphi
\end{array}
}\right)\,,\quad r\in[0,d]\,, \; \varphi\in[0,2\pi)\,.
\end{align*}
The normal vectors of the surfaces \begin{align*}
\overline{S}_1(z,\varphi)&=\left({ \begin{array}{c}
x_0+\sqrt{\frac{b-z}{c}}\,\cos \varphi\\
y_0+\sqrt{\frac{b-z}{c}}\,\sin \varphi\\
z
\end{array}
}\right)\,,\quad z \in \big[\tfrac{a+b}{2},b\big],\; \varphi \in [0, 2\pi]\,,\\
\overline{S}_2(z,\varphi)&=\left({ \begin{array}{c}
x_0+\sqrt{\frac{z-a}{c}}\,\cos \varphi\\
y_0+\sqrt{\frac{z-a}{c}}\,\sin \varphi\\
z
\end{array}
}\right)\,,\quad z \in \big[a,\tfrac{a+b}{2}\big],\; \varphi \in [0, 2\pi]\,,
\end{align*} are \begin{align*} \overline{N}_{S_1}(z,\varphi)&=\displaystyle\frac{\partial\overline{S}_1 }{\partial z}\times\frac{\partial\overline{S}_1 }{\partial \varphi}\\
&=\left|{\begin{array}{ccc}
\overline{e}_1 & \overline{e}_2 & \overline{e}_3\\
-\frac{1}{2\sqrt{c(b-z)}}\cos \varphi & -\frac{1}{2\sqrt{c(b-z)}}\sin\varphi & 1\\
-\sqrt{\frac{b-z}{c}}\sin\varphi & \sqrt{\frac{b-z}{c}}\cos\varphi & 0
\end{array}}\right|\\
&=-\sqrt{\frac{b-z}{c}}\sin\varphi\,\overline{e}_1-\sqrt{\frac{b-z}{c}}\cos\varphi\,\overline{e}_2-\frac{1}{2c}\,\overline{e}_3\,,\\ \overline{N}_{S_2}(z,\varphi)&=\displaystyle\frac{\partial\overline{S}_2 }{\partial z}\times\frac{\partial\overline{S}_2 }{\partial \varphi}\\
&=\left|{\begin{array}{ccc}
\overline{e}_1 & \overline{e}_2 & \overline{e}_3\\
\frac{1}{2\sqrt{c(z-a)}}\cos \varphi & \frac{1}{2\sqrt{c(z-a)}}\sin\varphi & 1\\
-\sqrt{\frac{z-a}{c}}\sin\varphi & \sqrt{\frac{z-a}{c}}\cos\varphi & 0
\end{array}}\right|\\
&=-\sqrt{\frac{z-a}{c}}\sin\varphi\,\overline{e}_1-\sqrt{\frac{z-a}{c}}\cos\varphi\,\overline{e}_2+\frac{1}{2c}\,\overline{e}_3\,,
\end{align*}
respectively. So, the area of $\partial{V}$ is equal to \begin{align*}
A(\partial{V})&=A(S_1)+A(S_2)\\
&=\displaystyle\mathop{\oiint}\limits_{S_1}{\big\|{\overline{N}_{S_1}(z,\varphi)}\big\|\,d(z,\varphi)}+\mathop{\oiint}\limits_{S_2}{\big\|{\overline{N}_{S_2}(z,\varphi)}\big\|\,d(z,\varphi)}\\
&=\bigintss_{\frac{a+b}{2}}^{b}\bigintss_{0}^{2\pi}{\sqrt{\Big(-\sqrt{\frac{b-z}{c}}\sin\varphi\Big)^2+\Big(-\sqrt{\frac{b-z}{c}}\cos\varphi\Big)^2+\Big(-\frac{1}{2c}\Big)^2}\;d\varphi\,dz}\,+\\
&\qquad\bigintss_a^{\frac{a+b}{2}}\bigintss_{0}^{2\pi}{\sqrt{\Big(-\sqrt{\frac{z-a}{c}}\sin\varphi\Big)^2+\Big(-\sqrt{\frac{z-a}{c}}\cos\varphi\Big)^2+\Big(\frac{1}{2c}\Big)^2}\;d\varphi\,dz}\,\\
&=\bigintsss_{\frac{a+b}{2}}^{b}\bigintsss_{0}^{2\pi}{\sqrt{\frac{b-z}{c}+
\frac{1}{4c^2}}\;d\varphi\,dz}+\bigintsss_a^{\frac{a+b}{2}}\bigintsss_{0}^{2\pi}{\sqrt{\frac{z-a}{c}+
\frac{1}{4c^2}}\;d\varphi\,dz}\\
&=\frac{\pi}{c}\int_{\frac{a+b}{2}}^{b}{\sqrt{4c(b-z)+1}\;dz}+\frac{\pi}{c}\int_a^{\frac{a+b}{2}}{\sqrt{4c(z-a)+
1}\;dz} \\
&=-\frac{\pi}{6c^2}\Big[\big(4c(b-z)+1\big)^{\frac{3}{2}}\Big]_{\frac{a+b}{2}}^{b}+\frac{\pi}{6c^2}\Big[\big(4c(z-a)+
1\big)^{\frac{3}{2}}\Big]_a^{\frac{a+b}{2}}
\\
&=-\frac{\pi}{6c^2}\big(1-(4c^2d^2+1)^{\frac{3}{2}}\big)+\frac{\pi}{6c^2}\big((4c^2d^2+1)^{\frac{3}{2}}-1\big)\\
&=\dfrac{\pi}{3c^2}\big((4c^2d^2+1)^{\frac{3}{2}}-1\big)\,.
\end{align*}

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