# What is the surface area of the solid created by revolving #f(x) = 3x, x in [2,5]# around the x axis?

By signing up, you agree to our Terms of Service and Privacy Policy

To find the surface area of the solid created by revolving the function ( f(x) = 3x ) on the interval ([2,5]) around the x-axis, you can use the formula for the surface area of revolution:

[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} , dx ]

First, calculate ( f'(x) ), then substitute it into the formula.

Given ( f(x) = 3x ), the derivative ( f'(x) ) is ( 3 ).

Substitute ( f(x) ) and ( f'(x) ) into the formula:

[ A = 2\pi \int_{2}^{5} 3x \sqrt{1 + 3^2} , dx ]

[ A = 2\pi \int_{2}^{5} 3x \sqrt{10} , dx ]

[ A = 6\pi \sqrt{10} \int_{2}^{5} x , dx ]

[ A = 6\pi \sqrt{10} \left[\frac{x^2}{2}\right]_{2}^{5} ]

[ A = 6\pi \sqrt{10} \left(\frac{5^2}{2} - \frac{2^2}{2}\right) ]

[ A = 6\pi \sqrt{10} \left(\frac{25}{2} - \frac{4}{2}\right) ]

[ A = 6\pi \sqrt{10} \left(\frac{21}{2}\right) ]

[ A = 63\pi \sqrt{10} ]

Therefore, the surface area of the solid is ( 63\pi \sqrt{10} ).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- What is a solution to the differential equation #dy/dx=x-y#?
- How do you solve for the equation #dy/dx=(3x^2)/(e^2y)# that satisfies the initial condition #f(0)=1/2#?
- What is the integrating factor for #0 = (3x^2 + 3y^2)dx + x(x^2 + 3y + 6y)dy#?
- What is a solution to the differential equation #dy/dx=-x/y# with the particular solution #y(1)=-sqrt2#?
- What is the surface area produced by rotating #f(x)=x/pi^2, x in [-3,3]# around the x-axis?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7