If the infinite curve $$ y=e^{-7 x}, x \geq 0 $$, is rotated about the $$ x $$-axis, find the area of the resulting surface. Need Help? Readit

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We start with the formula for the surface area of a curve $$ y=f(x) $$ rotated about the $$ x $$-axis:

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

Given $$ f(x) = e^{-7x} $$ for $$ x \ge 0 $$, we have:

$$
f'(x) = -7e^{-7x}.
$$

Thus, the integrand becomes:

$$
\sqrt{1+\left(-7e^{-7x}\right)^2} = \sqrt{1+49e^{-14x}}.
$$

Hence, the surface area is

$$
S = 2\pi \int_{0}^{\infty} e^{-7x}\sqrt{1+49e^{-14x}}\,dx.
$$

### Step 1. Use a substitution

Let

$$
u = e^{-7x}.
$$

Then

$$
\frac{du}{dx} = -7e^{-7x} = -7u \quad \Longrightarrow \quad dx = -\frac{du}{7u}.
$$

When $$ x=0 $$, we have $$ u = e^{0} = 1 $$, and as $$ x\to\infty $$, $$ u\to 0 $$.

Substituting these into the integral:

$$
S = 2\pi \int_{u=1}^{u=0} u \cdot \sqrt{1+49u^2} \left(-\frac{du}{7u}\right).
$$

Notice that the $$ u $$ in the numerator and denominator cancels, and we can reverse the limits when removing the negative sign:

$$
S = \frac{2\pi}{7} \int_{0}^{1} \sqrt{1+49u^2}\, du.
$$

### Step 2. Another substitution for the integral

Let

$$
t = 7u \quad \Longrightarrow \quad dt = 7\, du \quad \Longrightarrow \quad du = \frac{dt}{7}.
$$

When $$ u = 0 $$, $$ t = 0 $$ and when $$ u = 1 $$, $$ t = 7 $$. Then:

$$
\int_{0}^{1} \sqrt{1+49u^2}\, du = \int_{t=0}^{t=7} \sqrt{1+t^2}\cdot \frac{dt}{7} = \frac{1}{7} \int_{0}^{7} \sqrt{1+t^2}\, dt.
$$

Thus, the surface area becomes:

$$
S = \frac{2\pi}{7} \cdot \frac{1}{7}\int_{0}^{7} \sqrt{1+t^2}\, dt = \frac{2\pi}{49}\int_{0}^{7} \sqrt{1+t^2}\, dt.
$$

### Step 3. Evaluate the integral

The antiderivative of $$ \sqrt{1+t^2} $$ is given by:

$$
\int \sqrt{1+t^2}\, dt = \frac{t}{2}\sqrt{1+t^2} + \frac{1}{2}\ln\left|t+\sqrt{1+t^2}\right| + C.
$$

Evaluating from $$ t=0 $$ to $$ t=7 $$:

$$
\int_{0}^{7} \sqrt{1+t^2}\, dt = \left[ \frac{t}{2}\sqrt{1+t^2} + \frac{1}{2}\ln\left(t+\sqrt{1+t^2}\right) \right]_{0}^{7}.
$$

At $$ t=7 $$:

$$
\frac{7}{2}\sqrt{1+49} + \frac{1}{2}\ln\left(7+\sqrt{50}\right) = \frac{7}{2}\sqrt{50} + \frac{1}{2}\ln\left(7+\sqrt{50}\right).
$$

At $$ t=0 $$:

$$
0 + \frac{1}{2}\ln(0+1) = 0.
$$

Thus, the integral is:

$$
\int_{0}^{7} \sqrt{1+t^2}\, dt = \frac{7}{2}\sqrt{50} + \frac{1}{2}\ln\left(7+\sqrt{50}\right).
$$

### Step 4. Write the final answer

Substitute back into the expression for $$ S $$:

$$
S = \frac{2\pi}{49} \left[\frac{7}{2}\sqrt{50} + \frac{1}{2}\ln\left(7+\sqrt{50}\right)\right].
$$

Simplify by canceling the factor $$ \frac{2}{2} $$:

$$
S = \frac{\pi}{49}\left[7\sqrt{50} + \ln\left(7+\sqrt{50}\right)\right].
$$

Since $$ \sqrt{50} = 5\sqrt{2} $$, an equivalent expression is:

$$
S = \frac{\pi}{49}\left[35\sqrt{2} + \ln\left(7+5\sqrt{2}\right)\right].
$$

Thus, the area of the resulting surface is

$$
\boxed{S = \frac{\pi}{49}\left[35\sqrt{2} + \ln\left(7+5\sqrt{2}\right)\right]}.
$$
The area of the surface formed by rotating the curve $$ y = e^{-7x} $$ for $$ x \geq 0 $$ about the $$ x $$-axis is $$ \frac{\pi}{49}\left[35\sqrt{2} + \ln\left(7+5\sqrt{2}\right)\right] $$.

If the infinite curve \( y=e^{-7 x}, x \geq 0 \), is rotated about the \( x \)-axis, find the area of the resulting surface. Need Help? Readit

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