Find the volume of the solid formed by rotating about the $x$-axis

Question

Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis

I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.

$$\int_0^{0.7} \pi (e^x+2)^2dx$$

Answer 1

Your equation is correct. $$\begin{align*} A & = \pi \int_0^{0.7} (e^x + 2)^2\ dx = \pi \left( \int_0^{0.7} e^{2x}\ dx + 4 \int_0^{0.7} e^x\ dx + 4\cdot0.7 \right)\\ & = \pi \left( \frac12 \int_0^{1.4} e^x\ dx + 4 \int_0^{0.7} e^x\ dx + 2.8\right)\\ & = \pi \left( \frac12 (e^{1.4}-1) + 4(e^{0.7}-1) + 2.8 \right) \\ & \approx 26.3347\ldots \end{align*}$$

Answer 2

Your setup was all correct. However, you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also, bring out constants to LHS at the earliest, else that keeps tagging on till the end.

$$\begin{align*} A/\pi & = \int_0^{a} (e^x + c)^2\ dx = \int_0^{a} e^{2x}\ dx + 2c \int_0^{a} e^x\ dx + c^2 a\\ & = \frac12 ( e^{2a}-1) + 2c(e^{a}-1)\ + ac^2.\\ \end{align*}$$