volume of the solid generated when R is revolved about the x-axis

Key concept: This is a standard disk-method volume problem: the radius comes from the function value, and the volume is found with a definite integral.

Step by step

The region is bounded by $y=e^{x^2}$, the $x$-axis, the $y$-axis, and $x=2$ in the first quadrant. Revolving about the $x$-axis uses the disk method.

Step 1: Identify the radius

The radius of each disk is

$r=e^{x^2}$

Step 2: Write the area of a cross section

$A(x)=\pi r^2=\pi\left(e^{x^2}\right)^2=\pi e^{2x^2}$

Step 3: Set up the volume integral

The limits are from $x=0$ to $x=2$:

$V=\int_0^2 \pi e^{2x^2}\,dx$

Step 4: Evaluate with a calculator

This definite integral is evaluated numerically:

$V\approx 1266.125$

Answer

$\boxed{1266.125}$

So the correct choice is D.

Pitfall alert

A common mistake is squaring the exponent incorrectly, such as writing $e^{x^4}$ instead of $(e^{x^2})^2=e^{2x^2}$. Another error is using the washer method here; because the region touches the $x$-axis, disks are the correct cross sections.

Try different conditions

If the region were revolved about the $y$-axis instead, the setup would change completely and would likely require cylindrical shells or a different substitution. If the upper boundary were $y=e^x$ instead of $y=e^{x^2}$, the integral would become much simpler.

Further reading

disk method, volume of revolution, definite integral