The base of a certain solid is an equilateral triangle with altitude 8. Cross-sections perpendicular to the altitude are semicircles. Find the volume of the solid, using the formulaV=∫abA(x), with the left vertex of the triangle at the origin and the given altitude along the x-axis.

To find the volume of the solid, we'll use the formula \( V = \int_{a}^{b} A(x) \, dx \), where \( A(x) \) is the area of the cross-section at a particular value of \( x \).

### Step 1: Understanding the Geometry
The base of the solid is an equilateral triangle with an altitude of 8 units. This triangle lies in the \( xy \)-plane, with one vertex at the origin \((0,0)\), and the altitude extending along the \( x \)-axis. 

The equation for the lines bounding the triangle in the \( xy \)-plane can be derived as follows:
- The triangle's altitude extends from \( x = 0 \) to \( x = 8 \).
- For an equilateral triangle, the sides are symmetric about the altitude. The slopes of the sides are \( \pm \frac{\sqrt{3}}{3} \) because the altitude divides the base into two equal segments.

The equations of the lines can be written as:
- Left side: \( y = \frac{\sqrt{3}}{3} x \)
- Right side: \( y = -\frac{\sqrt{3}}{3} x + \frac{8\sqrt{3}}{3} \)

### Step 2: Cross-Sectional Area
Each cross-section perpendicular to the \( x \)-axis is a semicircle. The diameter of the semicircle at a given point \( x \) is the distance between the two sides of the triangle, which can be found as:

\[
\text{Diameter} = \text{Upper line} - \text{Lower line} = \left(-\frac{\sqrt{3}}{3} x + \frac{8\sqrt{3}}{3}\right) - \frac{\sqrt{3}}{3} x = \frac{8\sqrt{3}}{3} - \frac{2\sqrt{3}}{3} x
\]

Thus, the diameter at a point \( x \) is:

\[
D(x) = \frac{8\sqrt{3}}{3} - \frac{2\sqrt{3}}{3} x
\]

The radius \( r(x) \) of the semicircle is:

\[
r(x) = \frac{D(x)}{2} = \frac{4\sqrt{3}}{3} - \frac{\sqrt{3}}{3} x
\]

The area of the semicircular cross-section is:

\[
A(x) = \frac{1}{2} \pi r(x)^2 = \frac{1}{2} \pi \left(\frac{4\sqrt{3}}{3} - \frac{\sqrt{3}}{3} x \right)^2
\]

### Step 3: Set Up the Integral
We integrate \( A(x) \) from \( x = 0 \) to \( x = 8 \):

\[
V = \int_{0}^{8} \frac{1}{2} \pi \left(\frac{4\sqrt{3}}{3} - \frac{\sqrt{3}}{3} x \right)^2 dx
\]

### Step 4: Simplify and Integrate
Let's expand and simplify the integrand before integrating:

\[
V = \frac{\pi}{2} \int_{0}^{8} \left( \frac{48}{9} - \frac{24}{9} x + \frac{3}{9} x^2 \right) dx
\]

\[
V = \frac{\pi}{2} \int_{0}^{8} \left( \frac{16}{3} - \frac{8}{3} x + \frac{1}{3} x^2 \right) dx
\]

Now, integrate each term:

\[
V = \frac{\pi}{2} \left[ \frac{16}{3}x - \frac{4}{3}x^2 + \frac{1}{9}x^3 \right]_{0}^{8}
\]

Evaluate at \( x = 8 \):

\[
V = \frac{\pi}{2} \left[ \frac{16}{3}(8) - \frac{4}{3}(8)^2 + \frac{1}{9}(8)^3 \right]
\]

\[
V = \frac{\pi}{2} \left[ \frac{128}{3} - \frac{256}{3} + \frac{512}{9} \right]
\]

Simplifying further:

\[
V = \frac{\pi}{2} \left[ \frac{384 - 768 + 512}{9} \right] = \frac{\pi}{2} \left[ \frac{128}{9} \right] = \frac{64\pi}{9}
\]

### Final Answer:
The volume of the solid is:

\[
\boxed{\frac{64\pi}{9}}
\]

Do you want more details on any of these steps? Here are 5 related questions:

1. How do you find the equation of a line given the slope and a point?
2. Why do we use semicircles as the cross-section instead of full circles?
3. What is the significance of the equilateral triangle’s altitude in this problem?
4. Can this method be applied to solids with different base shapes?
5. How does the integral change if the cross-sections were different shapes?

**Tip:** When dealing with volumes of solids of known cross-section, always express the area of the cross-section as a function of the variable of integration before setting up the integral.

The base of a certain solid is an equilateral triangle with altitude 8. Cross-sections perpendicular to the altitude are semicircles. Find the volume of the solid, using the formula V=∫abA(x), with the left vertex of the triangle at the origin and the given altitude along the x-axis.

Explore how to calculate the volume of a solid whose base is an equilateral triangle with altitude 8 and cross-sections perpendicular to the altitude are semicircles. This advanced problem involves integral calculus and is suitable for Grades 11-12 students.

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