Answer: (a) \(\dfrac{dx}{dt}=2\sinh t\), and \(\dfrac{dy}{dt}=1-\sinh^2 t\).
(b)(i) \(A=\pi\displaystyle \int_0^1 \left(\dfrac32 t-\dfrac14\sinh 2t\right)(1+\cosh 2t)\,dt\).
(b)(ii) \(A=\pi\left(\dfrac34\sinh 2-\dfrac38\cosh 2-\dfrac1{16}(1+\cosh 2)^2+\dfrac{11}{8}\right)\).
(a) Differentiate the parametric equations with respect to \(t\).
Since \(x=2\cosh t\),
\(\dfrac{dx}{dt}=2\sinh t\).
Also \(y=\dfrac32 t-\dfrac14\sinh 2t\), so
\(\dfrac{dy}{dt}=\dfrac32-\dfrac14(2\cosh 2t)=\dfrac32-\dfrac12\cosh 2t\).
Using \(\cosh 2t=1+2\sinh^2 t\),
\(\dfrac{dy}{dt}=\dfrac32-\dfrac12(1+2\sinh^2 t)=1-\sinh^2 t\).
So \(\dfrac{dy}{dt}=1-\sinh^2 t\), as required.
(b)(i) For rotation about the \(x\)-axis,
\(A=2\pi\displaystyle \int_0^1 y\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\).
From part (a),
\(\dfrac{dx}{dt}=2\sinh t\), \(\dfrac{dy}{dt}=1-\sinh^2 t\).
Let \(s=\sinh t\). Then
\(\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2=4s^2+(1-s^2)^2\)
\(=4s^2+1-2s^2+s^4=s^4+2s^2+1=(s^2+1)^2\).
Since \(1+\sinh^2 t=\cosh^2 t\),
\(\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}=\cosh^2 t\).
Hence
\(A=2\pi\displaystyle \int_0^1 \left(\dfrac32 t-\dfrac14\sinh 2t\right)\cosh^2 t\,dt\).
Using \(\cosh^2 t=\dfrac12(1+\cosh 2t)\),
\(A=\pi\displaystyle \int_0^1 \left(\dfrac32 t-\dfrac14\sinh 2t\right)(1+\cosh 2t)\,dt\).
(b)(ii) Now evaluate
\(A=\pi\left[\dfrac32\int_0^1 t(1+\cosh 2t)\,dt-\dfrac14\int_0^1 \sinh 2t(1+\cosh 2t)\,dt\right]\).
For the second integral,
\(\int_0^1 \dfrac14\sinh 2t(1+\cosh 2t)\,dt=\left[\dfrac1{16}(1+\cosh 2t)^2\right]_0^1\)
\(=\dfrac1{16}(1+\cosh 2)^2-\dfrac14\).
For the first integral, integrate by parts with \(u=t\), \(dv=(1+\cosh 2t)dt\).
Then \(du=dt\) and \(v=t+\dfrac12\sinh 2t\), so
\(\dfrac32\int_0^1 t(1+\cosh 2t)\,dt=\dfrac32\left[t\left(\dfrac12\sinh 2t+t\right)\right]_0^1-\dfrac32\int_0^1 \left(\dfrac12\sinh 2t+t\right)dt\).
Thus
\(\dfrac32\int_0^1 t(1+\cosh 2t)\,dt=\dfrac32\left[t\left(\dfrac12\sinh 2t+t\right)-\dfrac14\cosh 2t-\dfrac12 t^2\right]_0^1\)
\(=\dfrac32\left(\dfrac12\sinh 2-\dfrac14\cosh 2+\dfrac34\right)\)
\(=\dfrac34\sinh 2-\dfrac38\cosh 2+\dfrac98\).
Therefore
\(A=\pi\left(\dfrac34\sinh 2-\dfrac38\cosh 2+\dfrac98-\left[\dfrac1{16}(1+\cosh 2)^2-\dfrac14\right]\right)\)
\(=\pi\left(\dfrac34\sinh 2-\dfrac38\cosh 2-\dfrac1{16}(1+\cosh 2)^2+\dfrac{11}{8}\right)\).
