IB Mathematics AHL 5.12 area of the region enclosed by a curve -AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Since the region is rotated about the \(y\)-axis, we integrate with respect to \(y\). A horizontal cross-section forms a circular disc of radius \(x\), so its area is \(\pi x^2\).
From
\(y=\dfrac{x^6}{65536}\),
we have
\(x^6=65536y\).
Therefore,
\(x^2=\sqrt[3]{65536y}\).
The volume of the bowl is
\(V=\pi\displaystyle\int_0^4x^2\,dy\).
\(V=\pi\displaystyle\int_0^4\sqrt[3]{65536y}\,dy\).
\(V=\pi\sqrt[3]{65536}\displaystyle\int_0^4y^{1/3}\,dy\).
\(V=\pi\sqrt[3]{65536}\left[\dfrac{3}{4}y^{4/3}\right]_0^4\).
\(V=192\pi\).
\(V=603.185\ldots\text{ cm}^3\).
✅ Answer: \(192\pi\text{ cm}^3\approx603\text{ cm}^3\)
(b)
Let the depth of the water be \(h\) centimetres. The volume of water is obtained by using \(h\) as the upper limit:
\(\pi\displaystyle\int_0^h\sqrt[3]{65536y}\,dy=250\).
Integrating gives
\(\pi\sqrt[3]{65536}\left[\dfrac{3}{4}y^{4/3}\right]_0^h=250\).
\(\dfrac{3\pi}{4}\sqrt[3]{65536}\,h^{4/3}=250\).
Therefore,
\(h=\left(\dfrac{1000}{3\pi\sqrt[3]{65536}}\right)^{3/4}\).
\(h=2.06622\ldots\).
The water therefore reaches approximately \(2.07\text{ cm}\) above the bottom of the bowl.
✅ Answer: \(2.07\text{ cm}\)