When a point is reflected across the x-axis the y coordinate becomes negative so when it's reflected across the x-axis the coordinates will be (x,-y).
When a point is reflected across the y-axis the x-coordinate becomes negative so when it's reflected across the y-axis the coordinates become (-x,y).
Answer:
Solution : Option B, or 9π
Step-by-step explanation:
We are given that y = x, x = 3, and y = 0.
Now assume we have a circle that models the given information. The radius will be x, so to determine the area of that circle we have πx². And knowing that x = 3 and y = 0, we have the following integral:

So our set up for solving this problem, would be such:

By solving this integral we receive our solution:
![\int _0^3x^2\pi dx,\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> \pi \cdot \int _0^3x^2dx\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1}\\=> \pi \left[\frac{x^{2+1}}{2+1}\right]^3_0\\=> \pi \left[\frac{x^3}{3}\right]^3_0\\\mathrm{Compute\:the\:boundaries}: \left[\frac{x^3}{3}\right]^3_0=9\\\mathrm{Substitute:9\pi }](https://tex.z-dn.net/?f=%5Cint%20_0%5E3x%5E2%5Cpi%20dx%2C%5C%5C%5Cmathrm%7BTake%5C%3Athe%5C%3Aconstant%5C%3Aout%7D%3A%5Cquad%20%5Cint%20a%5Ccdot%20f%5Cleft%28x%5Cright%29dx%3Da%5Ccdot%20%5Cint%20f%5Cleft%28x%5Cright%29dx%5C%5C%3D%3E%20%5Cpi%20%5Ccdot%20%5Cint%20_0%5E3x%5E2dx%5C%5C%5Cmathrm%7BApply%5C%3Athe%5C%3APower%5C%3ARule%7D%3A%5Cquad%20%5Cint%20x%5Eadx%3D%5Cfrac%7Bx%5E%7Ba%2B1%7D%7D%7Ba%2B1%7D%5C%5C%3D%3E%20%5Cpi%20%5Cleft%5B%5Cfrac%7Bx%5E%7B2%2B1%7D%7D%7B2%2B1%7D%5Cright%5D%5E3_0%5C%5C%3D%3E%20%5Cpi%20%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D%5E3_0%5C%5C%5Cmathrm%7BCompute%5C%3Athe%5C%3Aboundaries%7D%3A%20%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D%5E3_0%3D9%5C%5C%5Cmathrm%7BSubstitute%3A9%5Cpi%20%7D)
As you can tell our solution is option b, 9π. Hope that helps!
Just do what you would do if you multiply double digit numbers expect with an extra digit. then count how many numbers are behind the decimal and when your down put the decimal point as much to the left from how many numbers are originally behind the decimal.