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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

Sagot :

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

[tex]V = \pi \int\limits^a_b {x^2} \, dx[/tex]

Step(ii):-

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

The Volume of the solid

              [tex]V =\pi \int\limits^8_8 {(y^2+64)} \, dx[/tex]

               [tex]V = \pi (\frac{y^{3} }{3} + 64 y )_{-8} ^{8}[/tex]

              [tex]V = \pi (\frac{(8)^{3} }{3} + 64 (8) - (\frac{(-8)^{3} }{3} +64(-8) )[/tex]

             [tex]V = \pi (\frac{(8)^{3} }{3} + 64 (8) - (\frac{(-8)^{3} }{3} +64(-8) ) = \pi (\frac{1024}{3} + 1024)[/tex]

V = π(1365.33) cubic units