Find the volume of the solid formed when the region bounded by y = x² + 5, the x-axis, the y-axis and the line x = 5 is rotated 360° about the x-axis.

Step-by-step solution

1. 1

   Pin down the region and pick a method. The four boundaries give $0\le x\le 5$ (the $y$-axis is $x=0$) and $0\le y\le x^2+5$. Since $x^2+5>0$ everywhere on $[0,5]$, the curve never crosses the axis of rotation and the region touches that axis along its whole base — so each cross-section perpendicular to the $x$-axis is a full disk, not a washer. The disk method is the direct choice; the shell method would need the region split at $y=5$ and is more work.

2. 2

   Write the disk-method integral. A slice at position $x$ of thickness $dx$ sweeps a disk of radius $R(x)=y=x^2+5$ and area $\pi R(x)^2$, so

   $V=\pi\int_0^5 \left(x^2+5\right)^2 dx.$

3. 3

   Expand the square before integrating. There is no chain-rule shortcut here ($\int (x^2+5)^2dx \ne \tfrac{(x^2+5)^3}{3}$, because the derivative of the inside is not a constant), so multiply it out first:

   $\left(x^2+5\right)^2=x^4+10x^2+25.$

4. 4

   Integrate term by term. Each term is a simple power:

   $V=\pi\int_0^5\left(x^4+10x^2+25\right)dx=\pi\left[\frac{x^5}{5}+\frac{10x^3}{3}+25x\right]_0^5.$

5. 5

   Substitute the limits. At $x=5$: $\tfrac{5^5}{5}=625$, $\tfrac{10\cdot 125}{3}=\tfrac{1250}{3}$, and $25\cdot 5=125$. The lower limit contributes $0$. Adding over the common denominator $3$:

   $625+125+\frac{1250}{3}=\frac{1875+375+1250}{3}=\frac{3500}{3}.$

6. 6

   State the volume. Therefore

   $V=\frac{3500\pi}{3}\approx 3665.19\text{ cubic units}.$

   A numerical evaluation of $\pi\int_0^5(x^2+5)^2dx$ gives $3665.1914$, confirming the exact value.