Question

Find the volume of the of the solid enclosed by the paraboloid\(z = 2 + {x^2} + {\left( {y - 2} \right)^2}\)and the plane\(z = 1,x = 1,x = - 1,y = 0{\rm{ and }}y = 4\).

Short answer

Given:\(z = 2 + {x^2} + {\left( {y - 2} \right)^2}\)

To find: Volume

Step-by-step solution

Determining the limits

\( - 1 \le x \le 1,0 \le y \le 4,1 \le z \le 2 + {x^2} + {\left( {y - 2} \right)^2}\)

Choose\(z\)for the integrated. The order of integration is not important.

\(V = \int\limits_{ - 1}^1 {\int\limits_0^4 {\left( {2 + {x^2} + {{\left( {y - 2} \right)}^2}} \right) - 1dydx} } \)

Either simplify the integrated or break the integral into two parts and use substitution on one of them.

\(\begin{aligned}V &= \int\limits_{ - 1}^1 {\int\limits_0^4 {1 + {x^2} + {y^2} - 4ydydx} } \\ &= \int\limits_{ - 1}^1 {\int\limits_0^4 {5 + {x^2} + {y^2} - 4ydydx} } \end{aligned}\)

Integrating w.r.t \(y\).

\(\begin{aligned}V &= \int\limits_{ - 1}^1 {\left( {5y + {x^2}y + \frac{1}{3}{y^2} - \frac{4}{2}{y^2}} \right)dx} \\ &= \int\limits_{ - 1}^1 {\left( {20 + 4{x^2} + \frac{{64}}{3} - 32} \right)} - 0dx\\ &= \int\limits_{ - 1}^1 {\frac{{28}}{3} + 4{x^2}dx} \end{aligned}\)

Integrating w.r.t\(x\).

\(\begin{aligned}V &= \left( {\frac{{28}}{3}x + \frac{4}{3}{x^2}} \right)_{ - 1}^1\\ &= \left( {\frac{{28}}{3} + \frac{4}{3}} \right) - \left( {\frac{{ - 28}}{3} - \frac{4}{3}} \right)\\ &= \frac{{64}}{3}\end{aligned}\)

Therefore, volume of the solid is\(\frac{{64}}{3}\)cubic merits.