00:01
In this problem, we are asked to find out the surface area sa of the part of the sphere x squared plus y squared plus z squared equals to a squared that lies within the cylinder x squared plus y squared equals to a y and also it lies above the xy plane.
00:26
So here to calculate the surface integral we make use of the formula the double integral over the region.
00:33
D square root of xx squared plus zy squared plus 1 d a so here z equals to square root of a squared minus x squared minus y squared and this is obtained from the equation of the sphere let us partially differentiate this with respect to x we get negative x times a squared minus x squared minus y squared raised to the power negative 1 over 2 squaring on both sides we have xx squared to be equal to x squared x squared minus x squared minus y squared raised to the par negative 1.
01:15
Partially differentiating z with respect to y we get negative y times a squared minus x squared minus y squared raised to the power negative 1 over 2 and squaring on both the sides we get y squared times a squared minus x squared minus y squared the whole raise to the power negative 1 over 2.
01:36
So next we need to add both of these and add it to 1 that is we require xx squared plus z x squared plus zy squared plus 1 and this equals to x squared over a squared minus x squared minus y squared plus y squared over a squared minus x squared minus y squared plus one so now adding these we get x squared plus y squared plus a squared minus x squared minus y squared the whole divided by a squared minus x squared minus y squared.
02:15
So in the numerator x squared and negative x squared cancel out y squared and negative y squared cancel out.
02:21
So we are left with a squared over a squared minus x squared minus y squared.
02:28
So squaring on both sides we get square root of z x squared plus zy squared plus one to be equal to a over square root of a squared minus x squared minus y squared.
02:42
So next we are further given that the equation of the cylinder is x squared plus y squared is less than or equal to a y.
02:55
So here we're going to make use of the polar coordinates.
02:59
So x equals to r times cos of theta and y equals to r times sine of theta.
03:05
So we have r squared cost squared theta plus r squared sine squared theta is less than or equal to a times r times r times sine of theta.
03:16
So factoring out r squared we have r square times cost square teta plus sine squared theta which is less than or equal to a times r times sine of theta.
03:28
We know that cost square theta plus sine square teta is one so we get r squared is less than or equal to a times r times sine of theta.
03:36
Dividing by r on both sides we have r is less than or equal to a time sine of theta...