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
[tex] 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 [/tex]
