Edexcel IAL January 2022 P4 (WMA14/01) Q7 Volumes of Revolution, Integration by Parts
Figure 3 shows the design of a doorknob.
The shape of the doorknob is formed by rotating the curve shown in Figure 4 through \(360^\circ\) about the x-axis, where the units are centimetres.
The equation of the curve is given by
The shape of the doorknob is formed by rotating the curve shown in Figure 4 through \(360^\circ\) about the x-axis, where the units are centimetres.
The equation of the curve is given by
\(f(x) = \frac{1}{4} (4 – x)e^x \quad 0 \leq x \leq 4\)
(a) Show that the volume, \(Vcm^3\), of the doorknob is given by
\(V = k \int_0^4 (x^2 – 8x + 16) e^{2x} dx\)
where \(k\) is a constant to be found.(3)
(b) Hence, find the exact value of the volume of the doorknob.
Give your answer in the form \(p\pi(e^q + r)cm^3\) where \(p\), \(q\) and \(r\) are simplified rational numbers to be found.(5)
Give your answer in the form \(p\pi(e^q + r)cm^3\) where \(p\), \(q\) and \(r\) are simplified rational numbers to be found.(5)
Solution to Part a: Volume of Revolution
Key Concepts Used:
- Volume of Revolution Formula: \( V = \pi \int_{a}^{b} y^2 dx \) for rotation about the x-axis.
- Algebraic Simplification: Expand and simplify \( y^2 \).
- Constant Factorization: Factor out constants from the integral.
Step-by-Step Solution:
Given the curve: \(y = \frac{1}{4} (4 – x) e^{x}\)
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Compute \( y^2 \):
\(y^2 = \left( \frac{1}{4} (4 – x) e^{x} \right)^2 = \frac{1}{16} (4 – x)^2 e^{2x}\)
Expand \((4 – x)^2\):
\((4 – x)^2 = 16 – 8x + x^2\)So:
\(y^2 = \frac{1}{16} (x^2 – 8x + 16) e^{2x}\) -
Set Up the Volume Integral:
\(V = \pi \int_{0}^{4} y^2 dx = \pi \int_{0}^{4} \frac{1}{16} (x^2 – 8x + 16) e^{2x} dx\)
Factor out the constant:
\(V = \frac{\pi}{16} \int_{0}^{4} (x^2 – 8x + 16) e^{2x} dx\)Thus, the volume is given by:
\(V = k \int_{0}^{4} (x^2 – 8x + 16) e^{2x} dx\)
Final Answer:
\(k = \frac{\pi}{16}\)
Solution to Part b: Exact Volume of Revolution
Key Concepts Used:
- Integration by Parts: Apply to \(\int (x^2 – 8x + 16)e^{2x} dx\).
- Repeated Application: Use integration by parts twice to handle the polynomial.
- Definite Integration: Evaluate the antiderivative at the bounds \(x = 0\) and \(x = 4\).
- Exponential Integration: \(\int e^{2x} dx = \frac{1}{2} e^{2x} + C\).
Step-by-Step Solution:
We need to compute: \(I = \int_{0}^{4} (x^2 – 8x + 16)e^{2x} dx\)
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First Integration by Parts: Let:
\(u = x^2 – 8x + 16 \Rightarrow du = (2x – 8) dx\)
\(dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x}\)Apply:
\(I = \left[ \frac{1}{2} (x^2 – 8x + 16)e^{2x} \right]_{0}^{4} – \int_{0}^{4} \frac{1}{2} e^{2x} (2x – 8) dx\)Simplify:
\(I = \frac{1}{2} [(x^2 – 8x + 16)e^{2x}]_{0}^{4} – \int_{0}^{4} (x – 4)e^{2x} dx\) -
Second Integration for \(\int (x – 4)e^{2x} dx\): Let:
\(u = x – 4 \Rightarrow du = dx\)
\(dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x}\)Apply:
\(\int (x – 4)e^{2x} dx = \frac{1}{2} (x – 4)e^{2x} – \int \frac{1}{2} e^{2x} dx = \frac{1}{2} (x – 4)e^{2x} – \frac{1}{4} e^{2x}\) -
Substitute Back into \( I \):
\(I = \frac{1}{2} [(x^2 – 8x + 16)e^{2x}]_0^4 – \left[ \frac{1}{2}(x – 4)e^{2x} – \frac{1}{4}e^{2x} \right]_0^4\)
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Evaluate at \( x = 4 \):
\( x^2 – 8x + 16 = 16 – 32 + 16 = 0 \)
\((x – 4) = 0 \)
So:
\(\frac{1}{2}(0)e^8 – \left( \frac{1}{2}(0)e^8 – \frac{1}{4}e^8 \right) = 0 – (0 – \frac{1}{4}e^8) = \frac{1}{4}e^8\) -
Evaluate at \( x = 0 \):
\( x^2 – 8x + 16 = 0 – 0 + 16 = 16 \)
\((x – 4) = -4 \)
So:
\(\frac{1}{2}(16)e^0 – \left( \frac{1}{2}(-4)e^0 – \frac{1}{4}e^0 \right) = 8 – \left( -2 – \frac{1}{4} \right) = 8 + 2 + \frac{1}{4} = \frac{41}{4}\) -
Therefore:
\(I = \left( \frac{1}{4}e^8 \right) – \left( \frac{41}{4} \right) = \frac{1}{4}(e^8 – 41)\)
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Now Compute Volume:
\(V = \frac{\pi}{16} \cdot I = \frac{\pi}{16} \cdot \frac{1}{4}(e^8 – 41) = \frac{\pi}{64}(e^8 – 41)\)
So, \( p = \frac{1}{64}, q = 8, r = -41 \).
Final Answer
\(\frac{\pi}{64}(e^8 – 41)\)