Need help with high pass filter

Well here is my guess since impedance for a cap is 1/wc where w =2*PI*f where f is the frequency in Hz. looks like 33,000= 1/(2*PI*130*c) or c= 1/(2*PI*130*33,000) * 1,000,000 in uF = .0362uF How did I do guys (gals) ?

Thank you for your help, Joegio. But how do I incorporate the resistor into the formula (e.g. resistor of 15k ohm)? Is that just added to the amplifier's resistance (33,000 + 15,000), changing the equation to C= 1/(2*Pi*130*48,000)? Or is there another formula to incorporate the resistor? I know that this can be done without the resistor, connecting the ground directly, but based on a Stereophile prototype (Corey Greenberg Vol. 17 No. 1) who incorporated a resistor as well, there "must" be some advantage to that. Does anybody know?

Well I'm not sure about the resistor unless its to create a filter of a higher order. Anyway I took one of the Vandersteen filters apart and found two caps in parallel which sum the capacitance, but no resistor. If the resistor is in series than I think you would add it to the capacitive impedance, if in paralel than its 1/Z = 1/Xc + 1/Xr where Xc is capacative reactance and Xr is resistance reactance. This is where a spreadsheet comes in handy. Again I'm no expert, but I think this is how it goes.