Math Problem Statement

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.

Solution

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

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 xyxy-plane, with one vertex at the origin (0,0)(0,0), and the altitude extending along the xx-axis.

The equation for the lines bounding the triangle in the xyxy-plane can be derived as follows:

  • The triangle's altitude extends from x=0x = 0 to x=8x = 8.
  • For an equilateral triangle, the sides are symmetric about the altitude. The slopes of the sides are ±33\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=33xy = \frac{\sqrt{3}}{3} x
  • Right side: y=33x+833y = -\frac{\sqrt{3}}{3} x + \frac{8\sqrt{3}}{3}

Step 2: Cross-Sectional Area

Each cross-section perpendicular to the xx-axis is a semicircle. The diameter of the semicircle at a given point xx is the distance between the two sides of the triangle, which can be found as:

Diameter=Upper lineLower line=(33x+833)33x=833233x\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 xx is:

D(x)=833233xD(x) = \frac{8\sqrt{3}}{3} - \frac{2\sqrt{3}}{3} x

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

r(x)=D(x)2=43333xr(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)=12πr(x)2=12π(43333x)2A(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)A(x) from x=0x = 0 to x=8x = 8:

V=0812π(43333x)2dxV = \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=π208(489249x+39x2)dxV = \frac{\pi}{2} \int_{0}^{8} \left( \frac{48}{9} - \frac{24}{9} x + \frac{3}{9} x^2 \right) dx

V=π208(16383x+13x2)dxV = \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=π2[163x43x2+19x3]08V = \frac{\pi}{2} \left[ \frac{16}{3}x - \frac{4}{3}x^2 + \frac{1}{9}x^3 \right]_{0}^{8}

Evaluate at x=8x = 8:

V=π2[163(8)43(8)2+19(8)3]V = \frac{\pi}{2} \left[ \frac{16}{3}(8) - \frac{4}{3}(8)^2 + \frac{1}{9}(8)^3 \right]

V=π2[12832563+5129]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:

64π9\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Integral Calculus
Solid Geometry

Formulas

Volume of a solid with known cross-section: V = ∫abA(x) dx
Area of a semicircle: A(x) = (1/2)πr^2
Equation of a line in the xy-plane

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12