Certainly! Let’s find the surface area of revolution for the curve (y = 7x + 6) rotated about the y-axis over the interval ([1, b]).
The formula for the surface area of revolution about the y-axis is given by:
[ \text{Surface Area} = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]
In our case:
The curve is (y = 7x + 6).
The derivative (\frac{dy}{dx}) is (49\cos(7x)).
Now let’s calculate the surface area:
First, find the interval limits:
We are given the interval as ([1, b]). Since we need to find the surface area from (x = 1), we’ll set (a = 1).
Next, compute the integral: [ \text{Surface Area} = 2\pi \int_{1}^{b} (7x + 6) \sqrt{1 + 2401\cos^2(7x)} , dx ]
Evaluate the integral over the given interval.
Unfortunately, I don’t have the exact value of (b) to complete the calculation. If you provide the upper limit of the interval, I can give you the precise surface area.
