Figure 3 shows a sketch of part of the curve C with equation $y = 3^x$ The point P lies on C and has coordinates (2, 9) - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 7

To find the volume of the solid generated by rotating the region R around the x-axis, we will use the disk method.

The volume V can be expressed as:

$V = \pi \int_a^b [f(x)]^2 dx$

Here, (f(x) = 3^x) and we need to determine the bounds a and b. Since we already know the x-coordinate of point P is 2, and point Q is at (2 - \frac{1}{\ln(3)}), we set:

• Lower bound: (a = 0) (y-axis)
• Upper bound: (b = 2 - \frac{1}{\ln(3)})

Now we compute the volume:

$V = \pi \int_0^{2 - \frac{1}{\ln(3)}} (3^x)^2 dx$

This simplifies to:

$V = \pi \int_0^{2 - \frac{1}{\ln(3)}} 9^x dx$

Now we integrate:

$\int 9^x dx = \frac{9^x}{\ln(9)} + C$

Evaluating the limits:

$V = \pi \left[ \frac{9^{2 - \frac{1}{\ln(3)}}}{\ln(9)} - \frac{9^0}{\ln(9)} \right]$

Thus,

$V = \pi \left[ \frac{9^{2 - \frac{1}{\ln(3)}} - 1}{\ln(9)} \right]$

This expression gives the exact volume of the solid generated.