In cylindrical coordinates
x = r cos θ y = r sin θ z = z x=r\cos\theta\qquad
y=r\sin\theta\qquad
z=z x = r cos θ y = r sin θ z = z
In these coordinates the equation, x 2 + y 2 = 2 a y x^2+y^2=2ay x 2 + y 2 = 2 a y
r 2 = 2 a r sin θ or r = 2 a sin θ r^2=2ar\sin\theta\qquad\text{or}\qquad r=2a\sin\theta r 2 = 2 a r sin θ or r = 2 a sin θ
That is, r = f ( θ ) r=f(\theta) r = f ( θ ) f ( θ ) = 2 a sin θ f(\theta) =2a\sin\theta f ( θ ) = 2 a sin θ r ( θ , z ) = x ( θ , z ) ı ^ + y ( θ , z ) ȷ ^ + z ( θ , z ) k ^ \mathbf{r}(\theta,z) = x(\theta,z)\,\hat{\boldsymbol{\imath}} + y(\theta,z)\,\hat{\boldsymbol{\jmath}} + z(\theta,z)\,\hat{\mathbf{k}} r ( θ , z ) = x ( θ , z ) ^ + y ( θ , z ) ^ + z ( θ , z ) k ^
x ( θ , z ) = f ( θ ) cos θ = 2 a sin θ cos θ = a sin 2 θ y ( θ , z ) = f ( θ ) sin θ = 2 a sin θ sin θ = a ( 1 − cos 2 θ ) z ( θ , z ) = z \begin{aligned}
x(\theta,z)&=f(\theta)\cos\theta=2a\sin\theta\cos\theta=a\sin2\theta\\
y(\theta,z)&=f(\theta)\sin\theta=2a\sin\theta\sin\theta=a(1-\cos2\theta)\\
z(\theta,z)&=z
\end{aligned} x ( θ , z ) y ( θ , z ) z ( θ , z ) = f ( θ ) cos θ = 2 a sin θ cos θ = a sin 2 θ = f ( θ ) sin θ = 2 a sin θ sin θ = a ( 1 − cos 2 θ ) = z
Under this parametrization,
∂ r ∂ θ = 2 a cos 2 θ ı ^ + 2 a sin 2 θ ȷ ^ ∂ r ∂ z = k ^ ⟹ ∂ r ∂ θ × ∂ r ∂ z = − 2 a cos 2 θ ȷ ^ + 2 a sin 2 θ ı ^ ⟹ d S = ∣ ∂ r ∂ θ × ∂ r ∂ z ∣ d θ d z = 2 a d θ d z \begin{aligned}
\frac{\partial \mathbf{r}}{\partial \theta}=2a\cos2\theta\,\hat{\boldsymbol{\imath}}+2a\sin2\theta\,\hat{\boldsymbol{\jmath}}\qquad
\frac{\partial \mathbf{r}}{\partial z}=\hat{\mathbf{k}} &\implies \frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial z}
=-2a\cos2\theta\,\hat{\boldsymbol{\jmath}}+2a\sin2\theta\,\hat{\boldsymbol{\imath}} \\
&\implies \mathrm{d}S=\left|\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial z}\right|\,
\mathrm{d}\theta\,\mathrm{d}z
=2a\,\mathrm{d}\theta\,\mathrm{d}z
\end{aligned} ∂ θ ∂ r = 2 a cos 2 θ ^ + 2 a sin 2 θ ^ ∂ z ∂ r = k ^ ⟹ ∂ θ ∂ r × ∂ z ∂ r = − 2 a cos 2 θ ^ + 2 a sin 2 θ ^ ⟹ d S = ∂ θ ∂ r × ∂ z ∂ r d θ d z = 2 a d θ d z
We still have to determine the limits of integration.
The figure on the left below provides a top view of the cylinder.
From it we see that 0 ≤ θ ≤ π 0\le\theta\le\pi 0 ≤ θ ≤ π z 2 = x 2 + y 2 = r 2 z^2=x^2+y^2=r^2 z 2 = x 2 + y 2 = r 2 z = ± r z=\pm r z = ± r r = 2 a sin θ r=2a\sin\theta r = 2 a sin θ z 2 = r 2 = 4 a 2 sin 2 θ z^2=r^2=4a^2\sin^2\theta z 2 = r 2 = 4 a 2 sin 2 θ θ \theta θ z z z − 2 a sin θ -2a\sin\theta − 2 a sin θ z = + 2 a sin θ z=+2a\sin\theta z = + 2 a sin θ θ \theta θ
A r e a = ∫ ∣ z ∣ ≤ 2 a sin θ 2 a d θ d z = 2 a ∫ 0 π d θ ∫ − 2 a sin θ 2 a sin θ d z = 8 a 2 ∫ 0 π d θ sin θ = 8 a 2 [ − cos θ ] 0 π = 16 a 2 \begin{aligned}
{\rm Area} &= \int_{|z|\le 2a\sin\theta}2a\,\mathrm{d}\theta\,\mathrm{d}z
=2a\int_0^\pi \!\!\mathrm{d}\theta\int_{-2a\sin\theta}^{2a\sin\theta} \!\!\!\mathrm{d}z
=8a^2\int_0^\pi \!\!\mathrm{d}\theta\ \sin\theta
=8a^2\Big[-\cos\theta\Big]_0^\pi \\
&=16a^2
\end{aligned} Area = ∫ ∣ z ∣ ≤ 2 a s i n θ 2 a d θ d z = 2 a ∫ 0 π d θ ∫ − 2 a s i n θ 2 a s i n θ d z = 8 a 2 ∫ 0 π d θ sin θ = 8 a 2 [ − cos θ ] 0 π = 16 a 2
