I have two functions and I am attempting to find the area enclosed by each when rotated around $y=1$. These said functions $y=x$ and $y=\sqrt{x}$.
Upon graphing this it seemed simple however I am not sure what to integrate. So far I have:
Intercept: $(1,1)$, Its bounds are between $0$ and $1$.
To find the right angled triangled area without considering $y=\sqrt{x}$, so I did: $$\int_{0}^1 (1-x)^2\pi \,dx =\frac{\pi}{3}$$ then the area enclosed by the square root, so I did: $$\int_{0}^1 (\sqrt{x}-x)^2\pi \,dx =\frac{\pi}{30}$$
Then I did $\frac{\pi}{3}-\frac{3\pi}{10}$ to get my answer of $\frac{\pi}{30}$ however I know the answer should be $\frac{\pi}{6}$
Can anyone shed some light? Thank yoU!
$\int_0^1 (\pi (1-x)^2-\pi(1-\sqrt{x})^2 )dx$
This is what I did:
Do bigger circle area minus smaller circle area like
$\pi \cdot R^2-\pi \cdot r^2$