finding volume of solid

Question

Suppose that a solid is formed in such a way that each cross section perpendicular to the x-axis, for $0 \le x \le 1$, is a disk, a diameter of which goes from the x-axis out to the curve $y = \sqrt{x}$.

Find the volume of the solid.

For this I use the disk formula. So $$\pi\displaystyle\int_0^1(\sqrt{x})^2\,dx.$$ When I do this, I get $\dfrac{\pi}5$. The answer is $\dfrac{\pi}8$. What am I doing wrong?

Answer 1

As I wrote in the comments, and as well as @Soke pointed out, you are told that the cross-sections perpendicular to the $x$-axis are disks where the diameter is from $y=\sqrt{x}$ to $y=0$.

• The diameter of this disk is $\sqrt{x}$.
• The radius of this disk is $\sqrt{x}/2$.
• The cross-sectional area is $\pi (\sqrt{x}/2)^2$.
• $V = \int \text{"Cross-sectional Area"}\,\ d\text{ "axis perp to cross section"}$

$$V = \pi \int_0^1 \left(\dfrac{\sqrt{x}}{2}\right)^2\, dx = \pi\int_0^1 \dfrac{x}{4}\, dx = \left.\pi\dfrac{x^2}8\right|_0^1 = \dfrac\pi8$$

Answer 2

You're doing two things wrong.

One, it is explicitly saying that the diameter of each shell is $\sqrt{x}$, so the radius would be $\sqrt{x} / 2$.

Next, you're integrating wrong. The integral you have ($ \pi \int \limits_0^1 (\sqrt{x})^2 \, dx$) should give $\pi / 2$ which, multiplied by the new $\frac{1}{4}$ from above, gives the correct answer.