The base of a certain solid is an equilateral triangle with altitude 9. 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 for the volume of a solid with known cross-sectional area:

\[
V = \int_a^b A(x) \, dx
\]

### Step 1: Determine the bounds of integration and base of the triangle
The base of the solid is an equilateral triangle with an altitude of 9. The altitude runs along the \(x\)-axis from \(x = 0\) to \(x = 9\). Thus, the bounds of integration are \(a = 0\) and \(b = 9\).

### Step 2: Determine the length of the base of the triangle as a function of \(x\)
Since the cross-sections are semicircles, we need to find the diameter of these semicircles at each point along the \(x\)-axis.

For an equilateral triangle, the relationship between the altitude \(h\) and the side length \(s\) is given by:

\[
h = \frac{\sqrt{3}}{2} s
\]

Therefore, the side length of the triangle is:

\[
s = \frac{2h}{\sqrt{3}} = \frac{2 \times 9}{\sqrt{3}} = 6\sqrt{3}
\]

The triangle is symmetric with respect to the altitude, so the distance from the \(x\)-axis to the boundary of the triangle at a given \(x\) can be derived by using similar triangles.

Let \(y(x)\) be the length from the center (altitude) to one of the edges of the triangle. The total base length \(2y(x)\) of the triangle at any \(x\) is:

\[
y(x) = \frac{s}{2} - \frac{x}{\sqrt{3}}
\]

Substituting \(s = 6\sqrt{3}\), we get:

\[
y(x) = 3\sqrt{3} - \frac{x}{\sqrt{3}} = \frac{9 - x}{\sqrt{3}}
\]

The full base of the triangle at any \(x\) is then:

\[
2y(x) = \frac{2(9 - x)}{\sqrt{3}}
\]

### Step 3: Determine the area of the semicircle cross-section
The diameter of the semicircle at position \(x\) is \(2y(x)\), so the radius \(r(x)\) is:

\[
r(x) = y(x) = \frac{9 - x}{\sqrt{3}}
\]

The area of a semicircle is given by:

\[
A(x) = \frac{1}{2} \pi r(x)^2
\]

Substituting \(r(x)\):

\[
A(x) = \frac{1}{2} \pi \left(\frac{9 - x}{\sqrt{3}}\right)^2 = \frac{\pi}{2} \cdot \frac{(9 - x)^2}{3} = \frac{\pi}{6} (9 - x)^2
\]

### Step 4: Integrate to find the volume
The volume \(V\) is the integral of \(A(x)\) from \(x = 0\) to \(x = 9\):

\[
V = \int_0^9 \frac{\pi}{6} (9 - x)^2 \, dx
\]

### Step 5: Compute the integral

Let's evaluate the integral:

\[
V = \frac{\pi}{6} \int_0^9 (9 - x)^2 \, dx
\]

To evaluate this integral, we expand the square:

\[
(9 - x)^2 = 81 - 18x + x^2
\]

Now integrate each term:

\[
V = \frac{\pi}{6} \left[ \int_0^9 81 \, dx - 18 \int_0^9 x \, dx + \int_0^9 x^2 \, dx \right]
\]

\[
V = \frac{\pi}{6} \left[ 81x \bigg|_0^9 - 18 \cdot \frac{x^2}{2} \bigg|_0^9 + \frac{x^3}{3} \bigg|_0^9 \right]
\]

Substitute the limits:

\[
V = \frac{\pi}{6} \left[ 81(9) - 18 \cdot \frac{81}{2} + \frac{729}{3} \right]
\]

Simplify each term:

\[
V = \frac{\pi}{6} \left[ 729 - 729 + 243 \right] = \frac{\pi}{6} \cdot 243
\]

Finally:

\[
V = \frac{243\pi}{6} = 40.5\pi
\]

### Final Answer
The volume of the solid is \( \boxed{40.5\pi} \) cubic units.

Would you like more details on any of these steps, or do you have any questions?

Here are some related questions:

1. How do you find the centroid of an equilateral triangle?
2. What are other examples of solids with known cross-sectional areas?
3. How does the formula for the volume of a solid of revolution differ from this method?
4. Can the method of slicing be used for irregular shapes?
5. What is the significance of using definite integrals in volume calculation?

**Tip:** When working with integrals of functions involving geometric shapes, always consider symmetries and properties that can simplify your calculations.

The base of a certain solid is an equilateral triangle with altitude 9. Cross-sections perpendicular to the altitude are semicircles. Find the volume of the solid, using the formula V=∫(a to b) A(x), with the left vertex of the triangle at the origin and the given altitude along the x-axis.

Calculate the volume of a solid whose base is an equilateral triangle with altitude 9, and whose cross-sections are semicircles perpendicular to the altitude. This problem involves key calculus concepts such as integration and geometric relationships, suitable for Grades 11-12.

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