Question

Find the area of the region bounded by the curves y= -x²/3 + 2/3x + 5 and if -x+5.

Find also the volume when the area is rotated completely about the x-axis.

Answer 1

Taking curve as $Y_{1}=\frac{-1}{3}x^2+$ $\frac{2}{3}x+5$

and line as $Y_{2}=-x+5$

They meet at (0,5) and (5,0)

Area between the graphs

$A=\int_{0}^5(\frac{-1}{3}x^2+\frac{2}{3}x+5)dx-\int_{0}^5(-x+5)dx$

$|-\frac{1}{9}x^3+\frac{2}{3}x^2+5x|_0^5$ $-|\frac{-x^2}{2}+5x|_0^5$

$=(\frac{-125}{9}+\frac{50}{3}+25)-(-12.5+25)$

=19.44 - 12.5

=6.94

Volume $V=\pi\int_{0}^5(\frac{-1}{3}x^2+\frac{2}{3}x+5)^2dx-\pi\int_{0}^5(-x+5)^2dx$

$\int_0^5(\frac{-x^2}{3}+\frac{2}{3}x+5)^2$ dx

=$\int_0^5($ $\frac{x^4}{9}-\frac{4x^3}{9}-\frac{26}9x^2+\frac{20x}{3}+25$)dx

=$|\frac{x^5}{45}-\frac{x^4}{9}-\frac{26x^3}{27}+\frac{10x^2}{3}+25x|_0^5$

=87.96

$\int_0^5(-x+5)^2dx$

$=\int_0^5(x^2-10x+25)dx$

$=|\frac{x^3}{3}-5x^2+25x)|_0^5$

=41.67

$\pi(87.96-41.67) =46.29\pi$

=145.42