`int_0^1(2pi(3 - y)(1 - y^2))dy` Each integral represents the volume of a solid. Describe the solid.

Wednesday, March 5, 2008

$\displaystyle{\int_{{0}}^{{1}}}{\left({2}\pi{\left({3}-{y}\right)}{\left({1}-{{y}}^{{2}}\right)}\right)}{\left.{d}{y}\right.}$ Each integral represents the volume of a solid. Describe the solid.

Take out the constant $\displaystyle{2}\pi$ , and rewrite the integral.

$\displaystyle{\int_{{0}}^{{1}}}{\left({2}\pi{\left({3}-{y}\right)}{\left({1}-{{y}}^{{2}}\right)}{\left.{d}{y}\right.}\right.}$

= 2 $\displaystyle\pi$ $\displaystyle{\int_{{0}}^{{1}}}{\left({3}-{y}\right)}{\left({1}-{{y}}^{{2}}\right)}{\left.{d}{y}\right.}$

$\displaystyle={2}\pi{\int_{{0}}^{{1}}}{\left({3}{\left({1}-{{y}}^{{2}}\right)}-{y}{\left({1}-{{y}}^{{2}}\right)}{\left.{d}{y}\right.}={2}\pi{\int_{{0}}^{{1}}}{\left({3}-{3}{{y}}^{{2}}-{y}+{{y}}^{{3}}\right)}{\left.{d}{y}\right.}\right.}$

$\displaystyle={2}\pi{\left({3}{y}-{3}\frac{{{y}}^{{3}}}{{3}}-\frac{{{y}}^{{2}}}{{2}}+\frac{{{y}}^{{4}}}{{4}}\right)}{\mid}{\left({0},{1}\right)}$

$\displaystyle{2}\pi{\left({3}{y}-{{y}}^{{3}}-\frac{{{y}}^{{2}}}{{2}}+\frac{{{y}}^{{4}}}{{4}}\right)}{\mid}{\left({0},{1}\right)}$

$\displaystyle={2}\pi{\left({3}-{1}-\frac{{1}}{{2}}+\frac{{1}}{{4}}-{0}\right)}$

$\displaystyle={2}\pi{\left({2}-\frac{{1}}{{2}}+\frac{{1}}{{4}}\right)}$

$\displaystyle={2}\pi\frac{{{\left({2}\cdot{4}-{2}+{1}\right)}}}{{4}}$

$\displaystyle={2}\pi\frac{{{8}-{1}}}{{4}}={2}\pi{\left(\frac{{7}}{{4}}\right)}={7}\frac{\pi}{{2}}$

Thus the volume of the required solid is 7 $\displaystyle\frac{\pi}{{2}}$

Blog

Wednesday, March 5, 2008

$\displaystyle{\int_{{0}}^{{1}}}{\left({2}\pi{\left({3}-{y}\right)}{\left({1}-{{y}}^{{2}}\right)}\right)}{\left.{d}{y}\right.}$ Each integral represents the volume of a solid. Describe the solid.

No comments:

Post a Comment

What was the device called which Faber had given Montag in order to communicate with him?