Find the area of the surface generated by revolving the curve $x = \sqrt{25 - y^2}$, $-2 \leq y \leq 2$ about the $y$-axis.

Neetesh Kumar | November 15, 2024

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This is the solution to DHW Calculus
Assignment: 7 Question Number 9
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Step-by-step solution:

To find the surface area of the surface generated by revolving the curve $x = f(y)$ about the $y$-axis from $y = -2$ to $y = 2$, we use the surface area formula:

$S = 2 \pi \int_{a}^{b} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy$

In this case:

• $x = f(y) = \sqrt{25 - y^2}$
• $a = -2$ and $b = 2$

Step 1: Differentiate $x = f(y)$ with respect to $y$

$\frac{dx}{dy} = \frac{d}{dy} \left( \sqrt{25 - y^2} \right)$

Using the chain rule:

$\frac{dx}{dy} = \frac{1}{2 \sqrt{25 - y^2}} \cdot (-2y) = -\frac{y}{\sqrt{25 - y^2}}$

Step 2: Substitute into the Surface Area Formula

Now we have $x = \sqrt{25 - y^2}$ and $\frac{dx}{dy} = -\frac{y}{\sqrt{25 - y^2}}$.

Substitute these into the formula:

$S = 2 \pi \int_{-2}^{2} \sqrt{25 - y^2} \sqrt{1 + \left( -\frac{y}{\sqrt{25 - y^2}} \right)^2} \, dy$

Step 3: Simplify the Integrand

Simplify inside the square root:

$S = 2 \pi \int_{-2}^{2} \sqrt{25 - y^2} \sqrt{1 + \frac{y^2}{25 - y^2}} \, dy$

Combine the terms in the square root:

$S = 2 \pi \int_{-2}^{2} \sqrt{25 - y^2} \sqrt{\frac{25 - y^2 + y^2}{25 - y^2}} \, dy$

$S = 2 \pi \int_{-2}^{2} \sqrt{25 - y^2} \sqrt{\frac{25}{25 - y^2}} \, dy$

$S = 2 \pi \int_{-2}^{2} \sqrt{25 - y^2} \cdot \frac{5}{\sqrt{25 - y^2}} \, dy$

The $\sqrt{25 - y^2}$ terms cancel:

$S = 2 \pi \int_{-2}^{2} 5 \, dy$

$S = 10 \pi \int_{-2}^{2} 1 \, dy$

Step 4: Evaluate the Integral

$S = 10 \pi \cdot \left[ y \right]_{-2}^{2}$

$S = 10 \pi \cdot (2 - (-2))$

$S = 10 \pi \cdot 4$

$S = 40 \pi$

Final Answer:

$S = \boxed{40 \pi}$

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