So I'm to find the volume of the solid formed by the area trapped between $y = x$ and $y = \sqrt{x}$, rotated about $y = 1$
The two curves cross at y = 0 and y = 1 so they will be the upper/lower intervals.
Since we are rotating about the y = 1 and not the x axis, the $y = \sqrt{x}$ curve is the inner curve and $y = x$ is the outer curve?
Hence the area of the cross section is $A(x) = \pi(x^2-\sqrt{x}^2)$ i.e. $\pi(outer ~curve^2-inner~curve^2)$ which simplifies to $A(x) = \pi(x^2 – x)$
So volume = $V(x) =\pi\int_{0}^{1}(x^2-x)dx$
= $[\frac{x^3}{3}-\frac{x^2}{2}]^{1}_{0}$
= $\frac{1}{3}-\frac{1}{2}-0$
= $-\frac{1}{6}$
Is this the correct method, and if I receive a negative answer from the integral, I simply take the absolute value? Since the answers say $\frac{1}{6}$ or have I made a mistake in the working somewhere and should have received a postive answer?
Best Answer
The curves need to be thought of as with respect to the line $y=1$. What's the distance between the point $(x,\sqrt{x})$ and the line $y=1$? What's the distance between $(x,x)$ and $y=1$? Now compute
$$A=\int_0^1 \pi\left((1-x)^2-(1-\sqrt{x})^2\right)dx.$$