calculus 2: volume of solid of revolution formed by rotation of region

Question

Find the integral for the volume of the solid of revolution formed by rotating the region $R$ bounded by the curves $y = 4+x^2,\;x=0,\;y=0,\;and\;x=1\;$ about the $x$-axis in terms of $dx $ and $ dy$

Here is the graph: https://www.desmos.com/calculator/rij8mhgszd

I am still learning this so I would appreciate details! Thanks.

Edit:

Is this correct for dy?: $$A=\pi(\sqrt{y-4})^2 \\V=\pi\int_0^1(y-4)dy=\\\pi\big(\frac{y^2}2-4y\big)\left.\right\vert_{0}^{1}=\\\pi\big(\frac12-4\big)-0=\boxed{-3\frac{1}{2}\pi}$$

Answer 1

Hint:

The volume of the solid of revolution formed by rotating the curve $y=f(x)$ between $x=a$ and $x=b$ about the $x$-as is given by

$$\pi\int_a^b(f(x))^2\,dx$$

Hence, the volume in your problem is given by $$\pi\int_0^1(4+x^2)^2\,dx$$

Can you work out this integral?

---

Worked out solution:

$$\pi\int_0^1(4+x^2)^2\,dx=\pi\int_0^116+8x^2+x^4\,dx=\pi\big(16x+\frac83x^2+\frac15x^5\big)\left.\right\vert_{0}^{1}=\pi\big(16+\frac83+\frac15\big)=\boxed{18\frac{13}{15}\pi}$$