Volume of Solid in Calculus

Question

Find the volume of the solid whose base is the region in the first quadrant bounded by $y=x^5, y=1$ and the $y$-axis and whose cross sections perpeddicular to the $x$ axis are semicircles

I am tutoring someone, and we keep getting the wrong answer, any help would be appreciated

Answer 1

The diameter of the semicircular slice is $1 - x^5$, and its area then is $\pi(1-x^5)^2/8.$

The curve $y=x^5$ intersects the line $y=1$ at $x=1$ in the first quadrant, so your limits of integration are $[0,1]$.

Then, your infinitesimal slice has volume $(\pi(1-x^5)^2/8) dx$, and the volume then is

$$V = \frac{\pi}{8} \int_0^1 (1-x^5)^2 dx = \left.\frac{\pi}{8}(x - \frac{1}{3}x^6 + \frac{1}{11}x^{11})\right|_0^1 = \frac{25 \pi}{264}.$$

Answer 2

Your integral is $$\int_0^1 f(x) dx$$ Where $f(x)$ is the area of the semicircle at $x$. At $x$, the radius of the semicircle is $\frac12(1-x^5)$, so $f(x)=\frac\pi2(\frac12(1-x^5))^2$

What's left is to solve $$\int_0^1 \frac\pi2\left(\frac12(1-x^5)\right)^2 dx=\frac{25\pi}{264}$$

Answer 3

It should be

$$ A = \pi \int_{0}^{1} y^{2/5}dy = \frac{5}{7} \pi.$$