IB Mathematics AHL 5.12 area of the region enclosed by a curve -AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Set inner and outer surfaces equal at \( x = p \)
\( \frac{1}{2}p^3 + 1 = (p-1)^4 \)
Rearrange: \( \frac{1}{2}p^3 – (p-1)^4 + 1 = 0 \)
Solve numerically or graphically, \( p \approx 2.91082\ldots \)
To 3 s.f.: \( p = 2.91 \) cm
Result:
\( 2.91 \) cm
(b)
Volume of liquid is rotation of inner surface about \( y \)-axis
Express \( x \) in terms of \( y \): \( y = \frac{1}{2}x^3 + 1 \)
\( \frac{1}{2}x^3 = y – 1 \)
\( x^3 = 2(y – 1) \)
\( x = \sqrt[3]{2(y – 1)} \)
Radius squared: \( x^2 = (2(y – 1))^{\frac{2}{3}} \)
Upper limit: \( y \) at \( x = p = 2.91 \), \( y = \frac{1}{2}(2.91)^3 + 1 \)
\( (2.91)^3 \approx 24.645 \), \( \frac{1}{2} \times 24.645 \approx 12.3225 \), \( y \approx 12.3225 + 1 = 13.3225 \)
Integral: \( V = \pi \int_{1}^{13.3225\ldots} (2(y – 1))^{\frac{2}{3}} \, dy \)
Approximate: \( V \approx 196.946\ldots \) cm³
To 3 s.f.: \( V = 197 \) cm³
Result:
\( 197 \) cm³
(c)
Volume between surfaces is rotation of outer minus inner
Express \( x \) in terms of \( y \) for outer: \( y = (x – 1)^4 \)
\( (x – 1)^4 = y \)
\( x – 1 = y^{\frac{1}{4}} \)
\( x = y^{\frac{1}{4}} + 1 \)
Radius squared: \( x^2 = (y^{\frac{1}{4}} + 1)^2 \)
Integral: \( V_{\text{outer}} = \pi \int_{0}^{13.3225\ldots} (y^{\frac{1}{4}} + 1)^2 \, dy \)
Approximate: \( V_{\text{outer}} \approx 271.87668\ldots \) cm³
Subtract inner volume: \( 271.87668\ldots – 196.946\ldots \approx 74.93033\ldots \) cm³
To 3 s.f.: \( V = 74.9 \) cm³
Accept any value rounding to 75 cm³
Result:
\( 75 \) cm³