Inspired by Michael Mendrin

$As\quad question\quad said\quad top\quad corner\quad on\quad curve\quad and\quad a\quad side\quad is\quad on\quad \\ \\ x-axis.\\ \\ So\quad we\quad can\quad say\quad that\quad y-coordinate\quad of\quad both\quad corner\quad will\quad \\ \\ be\quad +ve.\\ \\ let\quad top\quad side\quad is\quad on\quad y=r\quad (r\ge 0),corner\quad will\quad be\quad points\quad on\\ \\ which\quad this\quad cut\quad curve\quad so,\\ \\ \cfrac { x }{ 1+{ x }^{ 2 } } =r\quad \Rightarrow x=\cfrac { 1\pm \sqrt { 1-4{ r }^{ 2 } } }{ 2r } \quad where\quad 4{ r }^{ 2 }\le 1\quad or\quad r\le \cfrac { 1 }{ 2 } \\ \\ so\quad corners\quad are\quad \left( \cfrac { 1\pm \sqrt { 1-4{ r }^{ 2 } } }{ 2r } ,0 \right) \& \left( \cfrac { 1\pm \sqrt { 1-4{ r }^{ 2 } } }{ 2r } ,r \right) \\ \\ length\quad of\quad rectangle\quad along\quad x-axis\quad L={ x }_{ 2 }{ -x }_{ 1 }=\cfrac { \sqrt { 1-4{ r }^{ 2 } } }{ r } \\ \\ radius\quad of\quad cylinder\quad R=r\\ \\ \therefore \quad Volume(V)=\pi { R }^{ 2 }L=\pi r\sqrt { 1-4{ r }^{ 2 } } =\pi \sqrt { { r }^{ 2 }-4{ r }^{ 4 } } \\ \\ =\cfrac { \pi }{ 2 } \sqrt { { 4r }^{ 2 }\left( 1-{ 4r }^{ 2 } \right) } \\ \\ Now\quad by\quad AP-GP\quad or\quad diffrentiation\quad you\quad will\quad find\quad V\quad \\ \\ will\quad bemax\quad for\quad { 4r }^{ 2 }=\cfrac { 1 }{ 2 } \\ \\ \Rightarrow { V }_{ max }=\cfrac { \pi }{ 4 } \\ \\ Ans=\left\lfloor \cfrac { \pi }{ 4 } \times 1000 \right\rfloor =785$