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

Answer:

V  = 36Π cubic Centimetres

Step-by-step 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

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

                   $V =\pi (\frac{x^{3} }{3} -9 x)^{3} _{-3}$

                 $V= \pi ( \frac{27}{3} - 9(3) - (\frac{-27}{3} -9(-3))$

                V = π ( |-36| = 36Π cubic Centimetres

Final answer:-

The volume of the solid

  V = π ( |-36| = 36Π cubic Centimetres