Question

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

Answer 1

Answer with explanation:

The equation of two curves are

y=8 sin x

y=8 cos x

The point of intersection of two curves are

→8 sin x=8 cos x

sinx = cos x

$x=\frac{\pi}{4}\\\\y=8\sin\frac{\pi}{4}\\\\y=4\sqrt{2}$

When you will look between points , 0 and $x=\frac{\pi}{4}\\\\y=8\sin\frac{\pi}{4}\\\\y=4\sqrt{2}$,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 $\frac{\pi}{4}$.

Required volume of solid obtained

    $=\int\limits^{\frac{\pi}{4}}_0 {8 \sin x} \, dx \\\\=|-8\cos x|\left \{ {{x=\frac{\pi}{4}}} \atop {x=0}} \right.\\\\=8-4\sqrt{2}$

    =8-4√2 cubic units