![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 }](/tpl/images/0296/0308/77546.png)
![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=\\\\\\ ](/tpl/images/0296/0308/998dc.png)
![=\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]=\\\\\\](/tpl/images/0296/0308/b9b37.png)
![=\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]=\\\\\\](/tpl/images/0296/0308/02840.png)
![=\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}](/tpl/images/0296/0308/813d2.png)
99 point question, plus !
( answer genuinely, and do not answer just for points, if you do, y...
Mathematics, 07.10.2019 04:30 91miketaylor
99 point question, plus !
( answer genuinely, and do not answer just for points, if you do, your answer will be deleted, and those points will be taken
this is calculus not basic
9.) set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.
y=6, y=12-(x^2)/12
a • v = pi*[integral from 6(sqrt(2)) to -6(sqrt(2))] ((12-(x^2)/12)^2)-36 dx = 2016/5 (sqrt(2))pi
b • v = pi*[integral from 6(sqrt(2)) to -6(sqrt(2))] ((12-(x^2)/12)^2)-36 dx = 4032/5 (sqrt(2))pi
c • v = pi*[integral from 12 to -12] ((12-(x^2)/12)^2)-36 dx = 4032/5 (sqrt(2))pi
d • v = pi*[integral from 12 to -12] ((12-(x^2)/12)^2)-36 dx = 2016/5 (sqrt(2))pi
** show all of you work, if you do not, your answer will be deleted, and the points you earned will be taken
Answers: 2
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