Question

The region in the first quadrant bounded above by the line yequals=77​, below by the curve yequals=startroot 7 x endroot7x​, and on the left by the ​y-axis is revolved about the line yequals=77. find the volume of the resulting solid.

Answer 1

The differential of volume can be a disk of radius (7-√(7x)) and thickness dx, so is

... dV = π·r²·dx = π(7-√(7x))²·dx

Integrated over the region 0 ≤ x ≤ 7, this becomes

$V=\displaystyle \int_{0}^{7}{\pi\left(7-\sqrt{7x}\right)^{2}\,dx}\\\\=\pi\left[49x-\dfrac{28x\sqrt{7x}}{3}+\dfrac{7x^2}{2}\right]\limits_{0}^{7}\\\\=49\pi\left(7-\dfrac{28}{3}+\dfrac{7}{2}\right)\\\\V=343\cdot \dfrac{\pi}{6}\approx 179.59438$