Problem 36 - Calculus Early Transcendentals

Question

Show that the surface area of the frustum of a cone generated by revolving the line segment between \((a, g(a))\) and \((b, g(b))\) about the \(x\) -axis is \(\pi(g(b)+g(a)) \ell,\) for any linear function \(g(x)=c x+d\) that is positive on the interval \([a, b],\) where \(\ell\) is the slant height of the frustum.

Step-by-Step Solution

Answer

Question: Show that the surface area of the frustum of a cone generated by revolving the line segment between (a, g(a)) and (b, g(b)) about the x-axis is π(g(b)+g(a))ℓ, for any linear function g(x)=cx+d that is positive on the interval [a, b]. Answer: A = π(g(b)+g(a))ℓ

Step 1: Find the equation of the line segment and radius expressions

To find the equation of the line segment between \((a, g(a))\) and \((b, g(b))\), we substitute the given linear function: $$g(x) = cx + d$$ Now, we need to express the radius of the frustum at a certain point x as a function of x. For this, we will substitute x=a and x=b, and we get: $$R_1 = g(a) = ca + d$$ $$R_2 = g(b) = cb + d$$

Step 2: Find the slant height of the frustum

To find the slant height (denoted as \(\ell\)), we have to use the Pythagorean theorem. We know the height difference between the two points is \(g(b) - g(a) = (cb+d) - (ca+d) = c(b-a)\). Let's denote the horizontal distance as \(h = b - a\). Thus, the slant height is given by: $$\ell=\sqrt{h^2+(g(b)-g(a))^2}=\sqrt{(b-a)^2+c^2(b-a)^2}=\sqrt{1+c^2}(b-a)$$

Step 3: Calculate the surface area of the frustum

Using the formula to find the surface area of a frustum of a cone, we have: $$A = \pi(R_1 + R_2) \ell$$ Now substituting the expression we found above: $$A = \pi((ca+d)+(cb+d))\sqrt{1+c^2}(b-a)$$ Simplify further: $$A=\pi(c(a+b)+2d)\sqrt{1+c^2}(b-a)$$

Step 4: Substitute the linear function and its values

We substitute the linear function \(g(x)=cx+d\) and its values (a, b) into the final expression we derived in the previous step: $$ A =\pi((g(a)+g(b))\sqrt{1+c^2}(b-a) $$

Step 5: Verify that the expression matches the desired result

Now, comparing our final expression to the desired result: \(\pi(g(b)+g(a)) \ell,\) we can see that they match: $$ A =\pi(g(b)+g(a))\sqrt{1+c^2}(b-a) = \pi(g(b)+g(a))\ell $$ Thus, we have shown that the surface area of the frustum of a cone generated by revolving the line segment between \((a, g(a))\) and \((b, g(b))\) about the \(x\) -axis is \(\pi(g(b)+g(a)) \ell\), for any linear function \(g(x)=cx+d\) that is positive on the interval \([a, b]\).