Question

Integral represents the volume of a solid. Describe the solid \(\int_{\rm{0}}^{\rm{1}} {\rm{2}} {\rm{\pi (3 - y)}}\left( {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dy}}\).

Short answer

The volume of the given integral is \(x = 1 - {y^2},x = 0,0 \le y \le 1,\) about the line \({\rm{y = 3}}\).

Step-by-step solution

Volume of the solid axis \({\rm{x}}\)and \({\rm{y}}\).

When a curve is rotated above the y-axis, it is said to be "rotated above the y-axis."

The solid obtained has a volume of

\({\rm{V = }}\int_{\rm{a}}^{\rm{b}} {\rm{2}} {\rm{\pi xydx}}\)

When a curve is rotated above the x-axis.

The solid obtained has a volume of,

\({\rm{V = }}\int_{\rm{a}}^{\rm{b}} {\rm{2}} {\rm{\pi xydy}}\).

Solid of given Integral.

Given Integral,

` \({\rm{V = }}\int_{\rm{0}}^{\rm{1}} {\rm{2}} {\rm{\pi (3 - y)}}\left( {{\rm{1 - }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dy}}\)

The region bordered by is one possible interpretation of the integral,

\(x = 1 - {y^2},x = 0,0 \le y \le 1,\)about the line \({\rm{y = 3}}\).

As an alternative, \({\rm{x = 1 - }}{{\rm{y}}^{\rm{2}}}\) ,\({\rm{x = }}{{\rm{y}}^{\rm{2}}}{\rm{,x = 1}}\) the darkened area would be horizontally flipped from left to right.

A different option would be to \({\rm{x = 1 - }}{{\rm{y}}^{\rm{2}}}\)which is \({\rm{y = }}\sqrt {{\rm{1 - x}}} \)Only the upper part of the parabola to be seen. So would not need to specify \(0 \le y \le 1\)because that can be deduced from the range of \({\rm{y = }}\sqrt {{\rm{1 - x}}} \)

It can't be less than \({\rm{0}}\) since it's impossible.

Volume of solid Graph.

Therefore, the region bounded \(x = 1 - {y^2},x = 0,0 \le y \le 1,\) about the line \({\rm{y = 3}}\).