Answer:
A is correct
Step-by-step explanation:
The base of a solid in the region bounded by the two parabolas [tex]y^2=8x[/tex] and [tex]x^2 = 8y[/tex].
Two parabola equation:
[tex]y_1=\dfrac{x^2}{8}[/tex]
[tex]y_2=\sqrt{8x}[/tex]
Cross sections of the solid perpendicular to the x-axis are semicircles.
Diameter of semicircle [tex]=y_1-y_2=\sqrt{8x}-\dfrac{x^2}{8}[/tex]
Radius of semicircle (r)[tex]=\sqrt{2x}-\dfrac{x^2}{16}[/tex]
Thickness of solid [tex]=dx[/tex]
Range of [tex]x: 0\leq x \leq 8[/tex]
Volume of solid = Area of semicircle x Thickness
[tex]V=\int_0^8\dfrac{1}{2}\pi\left ( \sqrt{2x}-\dfrac{x^2}{16} \right )^2dx\\\\V=\int_0^8\dfrac{1}{2}\pi\left ( 2x+\frac{x^4}{256}-\frac{\sqrt{2}x^{3/2}}{8} \right )dx\\\\[/tex]
[tex]V=\dfrac{\pi}{2}\left (x^2+\dfrac{x^5}{1280}-\dfrac{2\sqrt{2}x^{5/2}}{40} \right )|_0^8[/tex]
[tex]V=\dfrac{\pi}{2}(\frac{576}{35}-0)[/tex]
[tex]V=\dfrac{288\pi}{35}[/tex]
Hence, A is correct
