Find the volume of a solid generated by revolving the region bounded by the given lines adn curves about the x-axis.
$\displaystyle y=\sqrt2x+3, y=0,x=0,x=1$
This is wrong answer.I'm getting an answer of 2pi, is this correct?
You have $\displaystyle {y_1} = \sqrt 2 x + 3,{\text{ }}{y_2} = 0,{\text{ }}0 \leqslant x \leqslant 1,{\text{ }}{V_x} = ?$
Then $\displaystyle {V_x} = \pi \int\limits_{x = a}^{x = b} {\left( {y_1^2 - y_2^2} \right)dx} = \pi \int\limits_0^1 {{{\left( {\sqrt 2 x + 3} \right)}^2}dx} =$
$\displaystyle = \pi \int\limits_0^1 {\left( {2{x^2} + 6\sqrt 2 x + 9} \right)dx} = \pi \left. {\left( {\frac{2}{3}{x^3} + 3\sqrt 2 {x^2} + 9x} \right)} \right|_0^1 =$
$\displaystyle = \pi \left( {\frac{2}{3} + 3\sqrt 2 + 9} \right) = \boxed{\frac{\pi }
{3}\left( {29 + 9\sqrt 2 } \right)} \approx 43.6973778{\text{ }}\left( {{\text{cubic units}}} \right).$