Math  /  Calculus

QuestionFind the area of the surface generated by revolving the given curve about the xx-axis 1.) y=7xy = 7x, 0x10 \le x \le 1

Studdy Solution

STEP 1

What is this asking? We need to find the area of the surface created by spinning a line around the x-axis. Watch out! Remember the formula for surface area of revolution and don't forget to carefully calculate the derivative!

STEP 2

1. Define the function and the interval.
2. Calculate the derivative.
3. Set up the surface area integral.
4. Evaluate the integral.

STEP 3

Alright, so we're given the function y=7xy = 7x, which is a simple straight line.
We're interested in the part of the line between x=0x = 0 and x=1x = 1.
So, our **interval** is from x=0x = 0 to x=1x = 1.

STEP 4

Now, we need to find the derivative of our function y=7xy = 7x with respect to xx.
The derivative, written as dydx\frac{dy}{dx}, tells us the slope of the line at any point.

STEP 5

For y=7xy = 7x, the derivative is simply dydx=7\frac{dy}{dx} = 7.
See? Super easy!
The slope of the line is a **constant value**, which is 77.

STEP 6

The formula for the surface area generated by revolving a curve y=f(x)y = f(x) around the x-axis from x=ax = a to x=bx = b is given by: S=2πaby1+(dydx)2dx S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx

STEP 7

Let's plug in our values.
We have y=7xy = 7x, dydx=7\frac{dy}{dx} = 7, a=0a = 0, and b=1b = 1.
Substituting these into the formula gives us: S=2π01(7x)1+(7)2dx S = 2\pi \int_0^1 (7x) \sqrt{1 + (7)^2} dx S=2π017x1+49dx S = 2\pi \int_0^1 7x \sqrt{1 + 49} dx S=2π017x50dx S = 2\pi \int_0^1 7x \sqrt{50} dx

STEP 8

We can simplify 50\sqrt{50} as 525\sqrt{2}.
So, we have: S=2π017x52dx S = 2\pi \int_0^1 7x \cdot 5\sqrt{2} dx S=702π01xdx S = 70\sqrt{2}\pi \int_0^1 x dx

STEP 9

Now, we just need to evaluate the integral: S=702π01xdx S = 70\sqrt{2}\pi \int_0^1 x dx S=702π[x22]01 S = 70\sqrt{2}\pi \left[ \frac{x^2}{2} \right]_0^1 S=702π(122022) S = 70\sqrt{2}\pi \left( \frac{1^2}{2} - \frac{0^2}{2} \right) S=702π(120) S = 70\sqrt{2}\pi \left( \frac{1}{2} - 0 \right) S=702π12 S = 70\sqrt{2}\pi \cdot \frac{1}{2} S=352π S = 35\sqrt{2}\pi

STEP 10

The area of the surface generated by revolving y=7xy = 7x about the x-axis from x=0x = 0 to x=1x = 1 is 352π35\sqrt{2}\pi.

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