First of all, we have to draw our region. It is the one over the line y = 6 and below the parabola y = 12 - x²/12.
It is easy to notice that when we draw any straight line, that connect our line and parabola, it will be vertical and perpendicular to the axis of revolution x. If so, we use disk method. But first, we have to calculate limits of integration x₁ and x₂ (the limit values x of our region, in that points line y = 6 and parabola y = 12 - x²/12 have the same value). We have:
[tex]12 - \dfrac{x^2}{12}=6\quad|\cdot 12\\\\144-x^2=72\\\\x^2=144-72\\\\x^2=72\quad|\sqrt{(\dots)}\\\\x_1=-\sqrt{72}\qquad\vee\qquad x_2=\sqrt{72}\\\\
x_1=-\sqrt{36\cdot2}\qquad\vee\qquad x_2=\sqrt{36\cdot2}\\\\
\boxed{x_1=-6\sqrt{2}\qquad\vee\qquad x_2=6\sqrt{2}}[/tex]
Now we can calculate the volume:
[tex]V=\pi\cdot\int\limits_{-6\sqrt{2}}^{6\sqrt{2}} \left[\left(12-\dfrac{x^2}{12}\right)^2-6^2\right]\,dx=\\\\\\\boxed{V=\pi\cdot\int\limits_{-6\sqrt{2}}^{6\sqrt{2}} \left[\left(12-\dfrac{x^2}{12}\right)^2-36\right]\,dx }[/tex]
Answer A or B (of course if we reverse limits of integration; in this case integral from positive number to negative one will be negative, and we know, that V ≥ 0).
[tex]V=\pi\cdot\int\limits_{-6\sqrt{2}}^{6\sqrt{2}} \left[\left(12-\dfrac{x^2}{12}\right)^2-36\right]\,dx=\\\\\\=\pi\cdot\int\limits_{-6\sqrt{2}}^{6\sqrt{2}} \left[12^2-2\cdot12\cdot\dfrac{x^2}{12}+\left(\dfrac{x^2}{12}\right)^2-36\right]\,dx=\\\\\\=
\pi\cdot\int\limits_{-6\sqrt{2}}^{6\sqrt{2}} \left[144-2x^2+\dfrac{x^4}{144}-36\right]\,dx=\\\\\\=
\pi\cdot\int\limits_{-6\sqrt{2}}^{6\sqrt{2}} \left[\dfrac{x^4}{144}-2x^2+108\right]\,dx=\\\\\\
[/tex]
[tex]=\pi\cdot\left[\int\limits_{-6\sqrt{2}}^{6\sqrt{2}} \dfrac{x^4}{144}\, dx-\int\limits_{-6\sqrt{2}}^{6\sqrt{2}}2x^2\, dx+\int\limits_{-6\sqrt{2}}^{6\sqrt{2}}108\,dx\right]=\\\\\\=
\pi\cdot\left[\dfrac{1}{144}\int\limits_{-6\sqrt{2}}^{6\sqrt{2}} x^4\, dx-2\int\limits_{-6\sqrt{2}}^{6\sqrt{2}}x^2\, dx+108\int\limits_{-6\sqrt{2}}^{6\sqrt{2}}\,dx\right]=\\\\\\[/tex]
[tex]=\pi\cdot\left[\dfrac{1}{144}\cdot\left[\dfrac{x^5}{5}\right]_{-6\sqrt{2}}^{6\sqrt{2}}-2\cdot\left[\dfrac{x^3}{3}\right]_{-6\sqrt{2}}^{6\sqrt{2}}+108\cdot\left[x\right]_{-6\sqrt{2}}^{6\sqrt{2}}\right]=\\\\\\=
\pi\cdot\Biggl[\dfrac{1}{144}\left(\dfrac{(6\sqrt{2})^5}{5}-\dfrac{(-6\sqrt{2})^5}{5}\right)-2\left(\dfrac{(6\sqrt{2})^3}{3}-\dfrac{(-6\sqrt{2})^3}{3}\right)+\\\\\\+108\left(6\sqrt{2}-\Big(-6\sqrt{2}\right)\Big)\Biggr]=\\\\\\[/tex]
[tex]=\pi\Biggl[\dfrac{1}{144}\cdot\dfrac{31104\sqrt{2}+31104\sqrt{2}}{5}-2\cdot\dfrac{432\sqrt{2}+432\sqrt{2}}{3}+108\cdot12\sqrt{2}\Biggr]=\\\\\\
=\pi\cdot\Biggl[\dfrac{432\sqrt{2}}{5}-576\sqrt{2}+1296\sqrt{2}\Biggr]=\\\\\\=
\pi\cdot\Biggl[\dfrac{432\sqrt{2}}{5}-\dfrac{2880\sqrt{2}}{5}+\dfrac{6480\sqrt{2}}{5}\Biggr]=\boxed{\dfrac{4032\sqrt{2}}{5}\pi}[/tex]
Answer B.
