What is the area of the right region bounded by $x^2 + y^2 =25$ and $x = -3$?
My attempt solution: I solved first for the area of the left region and subtracted it from the area of the circle, which is
$$ 25 \pi - \int_{-4}^4 -3 + \sqrt {25-y^2} \,dy $$ $$ 78.5 - 11.18 = 67.32 $$ So, the area of the right region is 67.32. But what is the solution wherein I will not use the equation of the area of the circle, and just use integration alone?
"But is there a solution wherein I will not use the equation of the area of the circle, and just use integration alone?"
Yes, there are a number of ways, but they will not be simpler than the method you used.
One way, is to divide the region of the circle to the right of $x=-3$ into two halves, the region above $y=0$ and the region below. By symmetry, they have the same area. You can then use integration to find one of those areas and multiply by 2. If you had a similar problem, but without symmetry, you could simply integrate to find the top part, integrate to find the bottom part, and add the two parts.
A second way, would be to convert your problem to polar coordinates. This would be ugly, because of the line $x=-3$, but would give you the correct answer without subtracting from the area of a circle, and using integration alone.
One of the ways would be to change the order of your integral and rewrite as -
$\displaystyle \int_{-3}^5 \int_{-{\sqrt {25-x^2}}}^{\sqrt {25-x^2}} dy \, dx$
