Question

Let r be the region in the first quadrant bounded by the graph y=8- x^ (3/2) Find the area of the region R . Find the volume of the solid generated when R is revolved about the x-axis

Answer 1

The boundaries:
x = 0,  y = 8;  y = 0,  √x³ = = 8,  x = 4
$A= \int\limits^4_0 {(8- x^{3/2}) } \, dx = \\ =8 x - 2 \sqrt{ x^{5} }/5= \\ 8*4-64/5 =19.2$
$V = \int\limits^4_0 { \pi (8 - x^{3/2}) } \, dx = \\ = \pi \int\limits^0_4 {(64 - 16 * 2 \sqrt{ x^{5} /5} + x^{4}/4) \, dx =$
= π ( 64 x - 16 * 2 *√x^5  + x^4 / 4 ) =
= π ( 320 - 1024/5 + 64 ) = 179.2 π