Question

Consider the region bounded by the curves y=x2 and y=9. Find the volume of the solid formed when this region is rotated around the x-axis.

Answer 1

V = 36Π cubic Centimetres

Explanation:

Step(i):-

Volume of the solid

The Volume of the solid formed by revolving the region bounded by the curve y = f(x) and rotated around the x-axis defined by

`V = \pi \int\limits^a_b {[f(x)]^(2) } \, d x`

Step(ii):-

Given two curves are y = x² and y = 9

The point of intersection of two curves

y = x²...(i)

and y = 9 ...(ii)

Equating both equations , we get

x² - 9 =0

⇒ x² - 3² =0

⇒ (x+3)(x-3) =0

⇒ x+3=0 and x-3=0

x = 3 ⇒ y = 9

x = -3 ⇒y = 9

The point of intersection ( -3,9) and (3,9)

Step(iii):-

`V = \pi \int\limits^a_b {[(f(x)]^(2) } \, -[g(x)]^(2))d x`

The limits x- varies from -3 to 3

`V =\pi (\int\limits^3_3 {x^(2) } \, dx +\int\limits^3_3{9} \, dx`

`V =\pi ((x^(3) )/(3) -9 x)^(3) _(-3)`

`V= \pi ( (27)/(3) - 9(3) - ((-27)/(3) -9(-3))`

V = π ( |-36| = 36Π cubic Centimetres

Final answer:-

The volume of the solid

V = π ( |-36| = 36Π cubic Centimetres