Question
The curve $y = 4\ln(x – 2)$ for $0 \leq y \leq 4$ is rotated $360^\circ$ about the $y$-axis to form a solid of revolution.
Find the volume of the solid formed.
▶️Answer/Explanation
Detailed Solution
Step 1: Understanding the curve and the given bounds
The curve is given as:
\[
y = 4 \ln(x – 2)
\]
We’re restricted to the region where \( 0 \leq y \leq 4 \), and the curve is rotated 360 degrees about the \( y \)-axis to form a solid of revolution.
First, let’s determine the corresponding \( x \)-values for the given \( y \)-bounds.
– When \( y = 0 \):
\[
0 = 4 \ln(x – 2)
\]
\[
\ln(x – 2) = 0
\]
\[
x – 2 = e^0 = 1
\]
\[
x = 3
\]
– When \( y = 4 \):
\[
4 = 4 \ln(x – 2)
\]
\[
\ln(x – 2) = 1
\]
\[
x – 2 = e^1 = e
\]
\[
x = 2 + e
\]
So, the \( y \)-values from 0 to 4 correspond to \( x \)-values from \( x = 3 \) to \( x = 2 + e \). The domain of the function \( y = 4 \ln(x – 2) \) requires \( x – 2 > 0 \), i.e., \( x > 2 \), which is satisfied since \( x \) starts at 3 and increases.
Step 2: Decide on the method for finding the volume
Since the rotation is about the \( y \)-axis, there are two common approaches:
1. **Cylindrical Shells Method**: Integrate with respect to \( x \), using the formula:
\[
V = 2\pi \int_a^b x \cdot (\text{height of shell}) \, dx
\]
Here, the height of the shell at a given \( x \) is the \( y \)-value, i.e., \( y = 4 \ln(x – 2) \), and we integrate over \( x \) from 3 to \( 2 + e \).
2. Disk/Washer Method: Since the axis of rotation is the \( y \)-axis, we can express \( x \) as a function of \( y \) and integrate with respect to \( y \). The formula for the disk method (since the solid has no hole in this case) is:
\[
V = \pi \int_c^d [x(y)]^2 \, dy
\]
where \( x(y) \) is \( x \) expressed as a function of \( y \), and we integrate over \( y \) from 0 to 4.
The disk method looks more straightforward here because the bounds are given in terms of \( y \), and expressing \( x \) in terms of \( y \) aligns with the axis of rotation. Let’s proceed with the disk method, and we can use the shell method as a check if needed.
Step 3: Express \( x \) in terms of \( y \)
Start with the equation of the curve:
\[
y = 4 \ln(x – 2)
\]
Solve for \( x \):
\[
\ln(x – 2) = \frac{y}{4}
\]
\[
x – 2 = e^{y/4}
\]
\[
x = 2 + e^{y/4}
\]
So, \( x = 2 + e^{y/4} \). This gives us the radius of each disk at height \( y \). Since we’re rotating about the \( y \)-axis, the radius of the disk at a given \( y \) is \( x \), and the area of the disk is \( \pi x^2 \).
Step 4: Set up the integral using the disk method
The volume is:
\[
V = \pi \int_0^4 [x(y)]^2 \, dy
\]
Substitute \( x = 2 + e^{y/4} \):
\[
x^2 = \left(2 + e^{y/4}\right)^2 = 4 + 4e^{y/4} + e^{2(y/4)} = 4 + 4e^{y/4} + e^{y/2}
\]
So the integral becomes:
\[
V = \pi \int_0^4 \left(4 + 4e^{y/4} + e^{y/2}\right) \, dy
\]
Step 5: Compute the integral
Break it into three parts:
\[
V = \pi \left[ \int_0^4 4 \, dy + \int_0^4 4e^{y/4} \, dy + \int_0^4 e^{y/2} \, dy \right]
\]
First integral:
\[
\int_0^4 4 \, dy = 4y \Big|_0^4 = 4 \cdot 4 – 4 \cdot 0 = 16
\]
– **Second integral**:
\[
\int_0^4 4e^{y/4} \, dy
\]
Let \( u = \frac{y}{4} \), so \( y = 4u \), \( dy = 4 \, du \). When \( y = 0 \), \( u = 0 \); when \( y = 4 \), \( u = 1 \).
\[
\int_0^4 4e^{y/4} \, dy = 4 \int_0^1 e^u \cdot 4 \, du = 16 \int_0^1 e^u \, du
\]
\[
= 16 \left[ e^u \right]_0^1 = 16 (e^1 – e^0) = 16 (e – 1)
\]
Third integral:
\[
\int_0^4 e^{y/2} \, dy
\]
Let \( v = \frac{y}{2} \), so \( y = 2v \), \( dy = 2 \, dv \). When \( y = 0 \), \( v = 0 \); when \( y = 4 \), \( v = 2 \).
\[
\int_0^4 e^{y/2} \, dy = \int_0^2 e^v \cdot 2 \, dv = 2 \int_0^2 e^v \, dv
\]
\[
= 2 \left[ e^v \right]_0^2 = 2 (e^2 – e^0) = 2 (e^2 – 1)
\]
Combine:
\[
V = \pi \left[ 16 + 16(e – 1) + 2(e^2 – 1) \right]
\]
\[
= \pi \left[ 16 + 16e – 16 + 2e^2 – 2 \right]
\]
\[
= \pi \left[ 2e^2 + 16e – 2 \right]
\]
\[
= \pi \left[ 2(e^2 + 8e – 1) \right]
\]
\[
V = 2\pi (e^2 + 8e – 1) = 177 (approx)
\]
……………………Markscheme…………………………
Solution: –
attempts to express $x$ (or $x^2$) in terms of $y$
let $V$ be the volume of the solid formed
forms a definite integral of the form $\pi\int_c^d x^2 dy$ with their expression for $x^2$ in terms of $y$
$= 176.779…$
$= 177 (= 2\pi(e^2 + 8e – 1))$ cubic units)