Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 8 sin(x), y = 8 cos(x), 0 ≤ x ≤ π/4; about y = −1

Accepted Solution

Answer with explanation:The equation of two curves arey=8 sin xy=8 cos xThe point of intersection of two curves are→8 sin x=8 cos xsinx = cos x[tex]x=\frac{\pi}{4}\\\\y=8\sin\frac{\pi}{4}\\\\y=4\sqrt{2}[/tex]When you will look between points , 0 and [tex]x=\frac{\pi}{4}\\\\y=8\sin\frac{\pi}{4}\\\\y=4\sqrt{2}[/tex],the curve obtained is right angled triangle.Now, rotate the curve that is right angled triangle along the line , y= -1, to obtain a shape similar or resembling with Right triangular prism.Draw a circular disc in the right triangular prism , having radius =x,and small part on the triangle having length dx.dx varies from 0 to [tex]\frac{\pi}{4}[/tex].Required volume of solid obtained     [tex]=\int\limits^{\frac{\pi}{4}}_0 {8 \sin x} \, dx \\\\=|-8\cos x|\left \{ {{x=\frac{\pi}{4}}} \atop {x=0}} \right.\\\\=8-4\sqrt{2}[/tex]     =8-4√2 cubic units