Edexcel IAL June 2022 P4 (WMA14/01) Q8 Parametric Equation, Differentiation, Volume of Revolutions
Figure 2 shows the curve with equation
\( y = 10xe^{-\frac{1}{x}} \quad 0 \leq x \leq 10 \)
The finite region \( R \), shown shaded in Figure 2, is bounded by the curve, the x-axis and the line with equation \( x = 10 \)
The region \( R \) is rotated through \( 2\pi \) radians about the x-axis to form a solid of revolution.
(a) Show that the volume, \( V \), of this solid is given by
\( V = k \int_{0}^{10} x^2 e^{-x} dx \)
where \( k \) is a constant to be found.(2)
(b) Find \( \int x^2 e^{-x} dx \)(3)
Figure 3 represents an exercise weight formed by joining two of these solids together.
The exercise weight has mass 5kg and is 20 cm long.
Given that
\( \text{density} = \frac{\text{mass}}{\text{volume}} \times 100 \)
And using your answers to part (a) and part (b)
(c) find the density of this exercise weight. Give your answer in grams per cm\(^3\) to 3 significant figures.
Solution to Part a: Volume Integral Form
Key Concepts Used:
- Volume of Revolution: \( V = \pi \int_{a}^{b} y^2 dx \) for rotation about the x-axis.
- Algebraic Manipulation: Express \( y^2 \) and simplify.
Step-by-Step Solution:
-
Given curve:
\( y = 10xe^{-\frac{1}{2}x} \)
-
Compute \( y^2 \):
\( y^2 = (10xe^{-\frac{1}{2}x})^2 = 100x^2e^{-x} \)
since \( (e^{-\frac{1}{2}x})^2 = e^{-x} \). -
Volume integral:
\( V = \pi \int_{0}^{10} y^2 dx = \pi \int_{0}^{10} 100x^2e^{-x} dx = 100\pi \int_{0}^{10} x^2e^{-x} dx \)
-
Compare with given form:
\( V = k \int_{0}^{10} x^2e^{-x} dx \)
So, \( k = 100\pi \).
Final Answer:
\( 100\pi \)
Solution to Part b: Integrate \( \int x^2e^{-x} dx \)
Key Concepts Used:
- Integration by Parts: \( \int u dv = uv – \int v du \)
- Apply twice to reduce the power of \( x \)
Step-by-Step Solution:
-
Let
\( I = \int x^2 e^{-x} dx \)
(How to integrate \( x^2 e^{-x} \) ? Use integration by parts twice, reducing the power of \( x \) each time.)
-
First integration by parts: Let
\( u = x^2, \quad dv = e^{-x} dx \)
Then
\( du = 2x dx, \quad v = -e^{-x} \)
So,
\( I = -x^2 e^{-x} – \int (-e^{-x}) \cdot 2x dx = -x^2 e^{-x} + 2 \int xe^{-x} dx \) -
Second integration by parts for \( \int xe^{-x} dx \): Let
\( u = x, \quad dv = e^{-x} dx \)
Then
\( du = dx, \quad v = -e^{-x} \)
So,
\( \int xe^{-x} dx = -xe^{-x} – \int (-e^{-x}) dx = -xe^{-x} + \int e^{-x} dx = -xe^{-x} – e^{-x} \) -
Substitute back:
\( I = -x^2 e^{-x} + 2(-xe^{-x} – e^{-x}) = -x^2 e^{-x} – 2xe^{-x} – 2e^{-x} + C \)
Final Answer:
\( -x^2 e^{-x} – 2xe^{-x} – 2e^{-x} + C \)
Solution to Part c: Density of Exercise Weight
Key Concepts Used:
- Volume of one solid from part (a)
- Volume of two solids (exercise weight)
- Density formula: density = \(\frac{\text{mass}}{\text{volume}}\)
Step-by-Step Solution:
-
Volume of one solid:
From (a),
\( V_1 = 100\pi \int_{0}^{10} x^2 e^{-x} dx \)
Using the antiderivative from (b):
\( \int_{0}^{10} x^2 e^{-x} dx = [-x^2 e^{-x} – 2xe^{-x} – 2e^{-x}]_{0}^{10} \) -
Evaluate at limits:
○ At \( x = 10 \):
\( -100e^{-10} – 20e^{-10} – 2e^{-10} = -122e^{-10} \)
○ At \( x = 0 \):
\( 0 – 0 – 2e^{0} = -2 \)
So,
\( \int_{0}^{10} x^2 e^{-x} dx = (-122e^{-10}) – (-2) = 2 – 122e^{-10} \) -
Volume of one solid:
\( V_1 = 100\pi(2 – 122e^{-10}) \)
-
Volume of two solids (exercise weight):
\( V_{\text{total}} = 2 \times V_1 = 200\pi(2 – 122e^{-10}) \)
-
Mass: 5 kg = 5000 grams.
(Why convert mass to grams?
Density is asked in g/cm\(^3\), so mass must be in grams and volume in cm\(^3\).) -
Density:
\( \text{density} = \frac{5000}{200\pi(2 – 122e^{-10})} = \frac{25}{\pi(2 – 122e^{-10})} \)
-
Numerical evaluation:
– \( e^{-10} \approx 4.53999 \times 10^{-5} \)
– \( 122e^{-10} \approx 122 \times 4.53999 \times 10^{-5} \approx 0.005539 \)
– \( 2 – 0.005539 = 1.994461 \)
– \( \pi \times 1.994461 \approx 6.265 \)
– \( \frac{25}{6.265} \approx 3.99 \)
So, density = 3.99 g/cm\(^3\)
Final Answer:
3.99