The base of a solid is the region enclosed by the curve $\frac{x^4}{16}+\frac{y^4}{81}=1$

Q: How do you find the diameter of each cross section?

For a fixed x-value, the base curve gives the top and bottom y-values. The diameter of each semicircle is the vertical distance across the region, which is twice the positive y-value.

Q: What is the volume of the solid?

The volume is approximately 15.732 cubic units.

Question

The base of a solid is the region enclosed by the curve $\frac{x^4}{16}+\frac{y^4}{81}=1$ shown in the figure above. For the solid, each cross section perpendicular to the x-axis is a semicircle. What is the volume of the solid?
A 12.356
B 15.732
C 22.249

Acemy AI · Accepted Answer

**Key takeaway**: This is a cross-section volume problem: first find the diameter of each semicircle from the curve, then integrate the area.

For a fixed value of $x$, the base region satisfies
$$
\frac{x^4}{16}+\frac{y^4}{81}=1.
$$
So
$$
\frac{y^4}{81}=1-\frac{x^4}{16}
\quad\Rightarrow\quad
y^4=81\left(1-\frac{x^4}{16}\right).
$$
Taking the positive $y$ value for the top half of the region,
$$
y=3\left(1-rac{x^4}{16}\right)^{1/4}.
$$

Since each cross section perpendicular to the $x$-axis is a semicircle, the diameter is
$$
D(x)=2y=6\left(1-rac{x^4}{16}\right)^{1/4}.
$$
The area of a semicircle with diameter $D$ is
$$
A(x)=\frac{\pi}{8}D^2.
$$
Thus
$$
A(x)=\frac{\pi}{8}\cdot 36\left(1-rac{x^4}{16}\right)^{1/2}
=\frac{9\pi}{2}\sqrt{1-rac{x^4}{16}}.
$$

The region exists for $-2\le x\le 2$, so the volume is
$$
V=\int_{-2}^{2} \frac{9\pi}{2}\sqrt{1-rac{x^4}{16}}\,dx.
$$
Using symmetry,
$$
V=9\pi\int_0^2 \sqrt{1-rac{x^4}{16}}\,dx.
$$
Evaluating this integral numerically gives
$$
V\approx 15.732.
$$

So the volume is
$$
\boxed{15.732}
$$
which corresponds to choice **B**.

---
**Pitfalls the pros know** 👇
A frequent error is using the radius directly as the curve value $y$. For semicircular cross sections, the curve gives the half-height of the base region, so the diameter is $2y$, not $y$. Another common mistake is forgetting the symmetry of the region and integrating only over the positive side without doubling.

**What if the problem changes?**
If the cross sections were semicircles perpendicular to the $y$-axis instead, you would solve the equation for $x$ in terms of $y$, form the diameter as a function of $y$, and integrate with respect to $y$ over the appropriate interval.

`Tags`: cross-sectional area, semicircle, volume by slicing

Question

The base of a solid is the region enclosed by the curve $\frac{x^4}{16}+\frac{y^4}{81}=1$

Expert Verified Solution

FAQ

How do you find the diameter of each cross section?

What is the volume of the solid?