Question

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.

\({\rm{y = - }}{{\rm{x}}^{\rm{2}}}{\rm{ + 6x - 8,y = 0;}}\)about \({\rm{x}}\)axis.

Short answer

The volume of the curve is \({\rm{V = }}\frac{{{\rm{16\pi }}}}{{{\rm{15}}}}\).

Step-by-step solution

Curve height.

To solve for X, such that equations for the left and right parts of the parabola may be written

\({\rm{y = - }}{{\rm{x}}^{\rm{2}}}{\rm{ + 6x - 8}}\)

Subtract \({\rm{1}}\) from both sides of the equation.

\({\rm{y - 1 = - }}{{\rm{x}}^{\rm{2}}}{\rm{ + 6x - 9}}\)

Multiply all sides \({\rm{ - 1}}\).

\(\begin{aligned}{}{\rm{1 - y = }}{{\rm{x}}^{\rm{2}}}{\rm{ - 6x + 9}}\\{\rm{1 - y = (x - 3}}{{\rm{)}}^{\rm{2}}}\end{aligned}\)

Take the square root of,

\({\rm{ \pm }}\sqrt {{\rm{1 - y}}} {\rm{ = x - 3}}\)

Add \({\rm{3}}\),

\({\rm{3 \pm }}\sqrt {{\rm{1 - y}}} {\rm{ = x}}\)

Parabola left side\({\rm{3 - }}\sqrt {{\rm{1 - y}}} \)

Parabola right side\({\rm{3 + }}\sqrt {{\rm{1 - y}}} \)

So, the height is \({\rm{Height = (3 + }}\sqrt {{\rm{1 - y}}} {\rm{) - (3 - }}\sqrt {{\rm{1 - y}}} {\rm{) = 2}}\sqrt {{\rm{1 - y}}} \)

Volume of the solid graph.

Volume of the Solid.

\(\begin{aligned}{}{\rm{V = 2\pi }}\int_{\rm{0}}^{\rm{1}} {\rm{(}} {\rm{ Radius) (Height) dy}}\\{\rm{V = 4\pi }}\int_{\rm{0}}^{\rm{1}} {\rm{y}} \sqrt {{\rm{1 - y}}} {\rm{dy}}\end{aligned}\)

Insert the value \({\rm{1 - y = u}}\)and \({\rm{d y = d u}}\),

\(\begin{aligned}{}{\rm{V = - 4\pi }}\int_{\rm{1}}^{\rm{0}} {{\rm{(1 - u)}}} {{\rm{u}}^{{\rm{1/2}}}}{\rm{du}}\\{\rm{V = - 4\pi }}\int_{\rm{1}}^{\rm{0}} {{{\rm{u}}^{{\rm{1/2}}}}} {\rm{ - }}{{\rm{u}}^{{\rm{3/2}}}}{\rm{du}}\end{aligned}\)

\(\begin{aligned}{}{\rm{V = - 4\pi }}\left( {\frac{{\rm{2}}}{{\rm{3}}}{{\rm{u}}^{{\rm{3/2}}}}{\rm{ - }}\frac{{\rm{2}}}{{\rm{5}}}{{\rm{u}}^{{\rm{5/2}}}}} \right)_{\rm{1}}^{\rm{0}}\\{\rm{ = 0 - 4\pi }}\left( {\frac{{\rm{2}}}{{\rm{3}}}{\rm{ - }}\frac{{\rm{2}}}{{\rm{5}}}} \right)\\{\rm{ = }}\frac{{{\rm{16\pi }}}}{{{\rm{15}}}}\end{aligned}\)

Therefore, the volume of solid is \({\rm{V = }}\frac{{{\rm{16\pi }}}}{{{\rm{15}}}}\).