Answer
$$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^{2t})(t+1) \sqrt { t^2 + 2t + 2 } \space dt \approx 103.5999$$
Work Step by Step
1. Write the formula for the area of the surface obtained by rotating a curve about the $x$-axis.
$$S = \int_{\alpha}^{\beta} 2\pi y \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Calculate: $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = \frac{d(1 +te^t)}{dt} = e^t + te^t$
$\frac{dy}{dt} = \frac{d((t^2 + 1)e^t)}{dt} = 2te^t + (t^2 + 1)e^t $
3. Substitute the given values into the formula:
$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^t) \sqrt {(e^t + te^t)^2 + (2te^t + (t^2+1)e^t)^2} \space dt$
$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^t) \sqrt {(e^{2t} + 2e^tte^t+ t^2e^{2t}) + (4t^2e^{2t} + 2*(2te^t(t^2+1)e^t) + (t^2+1)^2e^{2t})} \space dt$
$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^t) \sqrt {e^{2t}(1 + 2t + t^2) + e^{2t} (4t^2 + 4t(t^2+1) + (t^2 + 1)^2)} \space dt$
$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^t) e^t \sqrt {(1 + 2t + t^2) + (2t + (t^2 + 1))^2} \space dt$
$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^t) e^t \sqrt {(t + 1)^2 + (t + 1)^4} \space dt$
$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^{2t})(t+1) \sqrt {1 + (t + 1)^2} \space dt$
$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^{2t})(t+1) \sqrt {1 + t^2 + 2t + 1} \space dt$
$S = \int_{0}^{1} 2\pi ((t^2 + 1)e^{2t})(t+1) \sqrt { t^2 + 2t + 2 } \space dt$
4. Write that integral on your calculator (you can use an online one).
The result is: $S \approx 103.5999$
