How do you find the area of the surface generated by rotating the curve about the y-axis #x=t+1, y=1/2t^2+t, 0

Question

How do you find the area of the surface generated by rotating the curve about the y-axis #x=t+1, y=1/2t^2+t, 0<=t<=2#?

Cameron Alexander · Accepted Answer

#= (2 pi)/3( 10 sqrt(10) - 2 sqrt(2)) " sq units"# #bbr(t) = << t+1, 1/2 t(t+2) >>, qquad 0<=t<=2# #ds = sqrt(dot x^2 + dot y^2) \ dt# #dS = ds * 2 pi x = 2 pi (t+1) sqrt((1)^2 + (t+1)^2) \ dt# #= 2 pi (t+1) sqrt(t^2 + 2t + 2) \ dt# #S =2 pi int_0^2 dt qquad (t+1) sqrt(t^2 + 2t + 2) # # =2 pi int_0^2 dt qquad d/dt ( 1/3(t^2 + 2t + 2)^(3/2)) # # =(2 pi)/3 [ (t^2 + 2t + 2)^(3/2)]_0^2 # # =(2 pi)/3 ( 10^(3/2) - 2^(3/2))# #= (2 pi)/3( 10 sqrt(10) - 2 sqrt(2)) approx 60.3 " sq units"#

Charles Anctil · Answer

undefined To find the area of the surface generated by rotating the curve about the y-axis, we can use the formula for the surface area of revolution, which is given by:

$$ A = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dx}{dt}\right)^2} dt $$

Given the curve $ x = t + 1 $ and $ y = \frac{1}{2}t^2 + t $, with $ 0 \leq t \leq 2 $, we need to find $ \frac{dx}{dt} $ and then substitute it into the formula.

First, calculate $ \frac{dx}{dt} $:

$$ \frac{dx}{dt} = \frac{d}{dt}(t + 1) = 1 $$

Now, we substitute $ \frac{dx}{dt} = 1 $ and $ y = \frac{1}{2}t^2 + t $ into the formula:

$$ A = \int_{0}^{2} 2\pi \left(\frac{1}{2}t^2 + t\right) \sqrt{1 + (1)^2} dt $$

Simplify the expression:

$$ A = \int_{0}^{2} 2\pi \left(\frac{1}{2}t^2 + t\right) \sqrt{2} dt $$

$$ A = \sqrt{2}\pi \int_{0}^{2} (t^2 + 2t) dt $$

Now integrate term by term:

$$ A = \sqrt{2}\pi \left[\frac{1}{3}t^3 + t^2\right]_{0}^{2} $$

$$ A = \sqrt{2}\pi \left[\frac{1}{3}(2)^3 + (2)^2 - \left(\frac{1}{3}(0)^3 + (0)^2\right)\right] $$

$$ A = \sqrt{2}\pi \left[\frac{8}{3} + 4\right] $$

$$ A = \sqrt{2}\pi \left[\frac{8}{3} + \frac{12}{3}\right] $$

$$ A = \sqrt{2}\pi \cdot \frac{20}{3} $$

$$ A = \frac{20\sqrt{2}\pi}{3} $$

Therefore, the area of the surface generated by rotating the curve about the y-axis is $ \frac{20\sqrt{2}\pi}{3} $.

How do you find the area of the surface generated by rotating the curve about the y-axis #x=t+1, y=1/2t^2+t, 0&lt;=t&lt;=2#?

How do you find the area of the surface generated by rotating the curve about the y-axis #x=t+1, y=1/2t^2+t, 0<=t<=2#?