A tangent of the curve $y=\sqrt{x}$ at point Q crosses the point P(-1,0). What is the volume of the solid of revolution when the area bounded by the line PQ, $y=\sqrt{x}$ and the x-axis is rotated through 2$\pi$ about the x-axis?
My approach:
Setting the x-coordinate of point Q as $a$, then Q($a$,$\sqrt{a}$)
The equation of the tangent can be written as $$y-\sqrt{a}={{1}\over{2\sqrt{a}}}(x-a)$$ Substituting $x=-1$ and $y=0$ into the equation, $$-\sqrt{a} = {{-1-a}\over{2\sqrt{a}}}$$ $$-2a = -1-a$$ therefore $a=1$
Now point Q(1,1) can be used to come up with an equation for the line PQ, which is $$y={{1}\over{2}}(x+1)$$ Then, the volume $V$ can be calculated as $$V=\pi\int^1_{-1}\left({{1}\over{2}}(x+1)\right)^2-(\sqrt{x})^2\ dx$$ $$=\pi\int^1_{-1}{{1}\over{4}}(x+1)^2-x\ dx$$ $$=\pi\int^1_{-1}{{1}\over{4}}(x^2+2x+1)-x\ dx$$ $$=\pi\int^1_{-1}{{1}\over{4}}x^2-{{1}\over{2}}x+{{1}\over{4}}\ dx$$ $$=\pi\left[{{1}\over{12}}x^3-{{1}\over{4}}x^2+{{1}\over{4}}x\right]^1_{-1}$$ which returns ${2\pi}\over{3}$.
Did I do this question correctly?
There is an error in your work insofar as the limits of integration on the $\sqrt{x}$-term is incorrect. This should be $x\in[0,1]$, as seen in the figure below. Thus we have
$$ \begin{align} V &=\pi\left(\int_{-1}^{1}\left(\frac{x+1}{2} \right)^2~dx -\int_0^1 \sqrt{x}^2~dx \right)\\ &=\pi\left(\frac{(x+1)^3}{3\cdot4} \biggr|_{-1}^1 + \frac{x^2}{2}\biggr|_0^1\right)\\ &=\frac{\pi}{6} \end{align} $$
I have verified this result numerically.