Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations and inequalities about the y-axis.x2 - y2 = 64, x \geq 0, y = -8, y = 8

Answer 1

Answer:

The Volume of the solid

V = π(1365.33) cubic units

Step-by-step explanation:

Step(I):-

Given curves are  x²- y² = 64 ,  x=0 , y=-8 and y=8

The given function   x = f(y)

                            x²- y² = 64

                            x² = y² +64

The Volume of the solid generating by revolving the region bounded by the graphs of the equations

$V = \pi \int\limits^a_b {x^2} \, dx$

Step(ii):-

the limits are  y = a = -8 and y = b=8

The Volume of the solid

              $V =\pi \int\limits^8_8 {(y^2+64)} \, dx$

               $V = \pi (\frac{y^{3} }{3} + 64 y )_{-8} ^{8}$

              $V = \pi (\frac{(8)^{3} }{3} + 64 (8) - (\frac{(-8)^{3} }{3} +64(-8) )$

             $V = \pi (\frac{(8)^{3} }{3} + 64 (8) - (\frac{(-8)^{3} }{3} +64(-8) ) = \pi (\frac{1024}{3} + 1024)$

V = π(1365.33) cubic units