Question

Compute the area of the surface formed by revolving about the $x$ -axis the cardioid $x = a (2 \cos t - \cos 2 t), y = a (2 \sin t - \sin 2 t) .$

Accepted Answer

Given;

The cardioid,

$x = a (2 \cos t - \cos 2 t), y = a (2 \sin t - \sin 2 t) .$

Here, we are going to find the area of the surface formed by revolving about the x-axis the cardioid.

Now, by differentiating both the equation with respect to $t$ .

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} a (2 \cos t - \cos 2 t) \Rightarrow \frac{d x}{d t} = a (2 \sin 2 t - 2 \sin t)$

$\Rightarrow \frac{d y}{d t} = \frac{d}{d t} a (2 \sin t - \sin 2 t) \Rightarrow \frac{d y}{d t} = a (2 \cos t - 2 \cos 2 t)$

As the cardioid is revolving about the $x$ -axis.

Then, Area:

$\Rightarrow A = 2 π \int_{t_{2}}^{t_{1}} y (t) \sqrt{{(\frac{dx}{dt})}^{2} + {(\frac{dy}{dt})}^{2}} dt \Rightarrow A = 2 π \int_{t_{2}}^{t_{1}} [a (sint - \sin 2 t)] \sqrt{{[a (2 \sin 2 t - 2 sint)]}^{2} + {[a (2 \cos t - 2 \cos 2 t)]}^{2}} dt \Rightarrow A = 2 π \int_{t_{2}}^{t_{1}} [a (sint - \sin 2 t)] \sqrt{[a^{2} {(2 \sin 2 t - 2 sint)}^{2}] + [a^{2} {(2 \cos t - 2 \cos 2 t)}^{2}]} dt \Rightarrow A = 2 π \int_{t_{2}}^{t_{1}} [a^{2} (sint - \sin 2 t)] \sqrt{4 \sin^{2} 2 t + 4 \sin^{2} t - 2 \times 2 \sin 2 t \times 2 sint + 4 \cos^{2} t + 4 \cos^{2} 2 t - 2 \times 2 \cos 2 t \times 2 cost} ∴ \sin^{2} θ + \cos^{2} θ = 0 \Rightarrow A = 2 π \int_{t_{2}}^{t_{1}} [a^{2} (sint - \sin 2 t)] \sqrt{4 \sin^{2} 2 t + 4 \cos^{2} 2 t + 4 \sin^{2} t + 4 \cos^{2} t - 8 . sint . \sin 2 t - 8 . cost . \cos 2 t} \Rightarrow A = 2 π \int_{t_{2}}^{t_{1}} [a^{2} (sint - \sin 2 t)] \sqrt{4 (\sin^{2} 2 t + \cos^{2} 2 t) + 4 (\sin^{2} t + \cos^{2} t) - 8 . sint . \sin 2 t - 8 . cost . \cos 2 t} \Rightarrow A = 2 π \int_{t_{2}}^{t_{1}} [a^{2} (sint - \sin 2 t)] \sqrt{4 + 4 - 8 . sint . \sin 2 t - 8 . cost . \cos 2 t} \Rightarrow A = 2 π \int_{t_{2}}^{t_{1}} [a^{2} (sint - \sin 2 t)] \sqrt{8 - 8 . sint . \sin 2 t - 8 . cost . \cos 2 t} \Rightarrow A = 4 \sqrt{2} π \int_{t_{2}}^{t_{1}} [a^{2} (sint - \sin 2 t)] \sqrt{1 - sint . \sin 2 t - cost . \cos 2 t}$

By calculating it we will get:

$\Rightarrow A = \frac{128}{5} π a^{2}$

Hence, The area of the surface is $\frac{128}{5} {πa}^{2}$ Square unit.

Compute the area of the surface formed by revolving about the $x$ -axis the cardioid $x = a (2 \cos t - \cos 2 t), y = a (2 \sin t - \sin 2 t) .$

Important Questions on Area under Curves

Learn and Prepare for any exam you want !

Compute the area of the surface formed by revolving about the x-axis the cardioid x=a2cost-cos2t, y=a2sint-sin2t.

Important Questions on Area under Curves

Learn and Prepare for any exam you want !

Compute the area of the surface formed by revolving about the $x$ -axis the cardioid $x = a (2 \cos t - \cos 2 t), y = a (2 \sin t - \sin 2 t) .$