00:01
Hi, in this question, given that the surface z equals y square plus 4x lying above the triangular region xy plane with the vertices 0 ,0 ,0 ,2 and 2 ,2, we need to find the surface area.
00:23
We know that the formula to find the surface area s equals double integral over square root of 1 plus dou f by dou x the whole square plus dou f by dou y the whole square into dx dy.
00:41
So, here this is the line x equals y and this is the point 0 ,2 and this is the point 2 ,2 and this is the point 2 ,0 and this is 0 ,0.
00:58
So, here first we need to find dou f by dou x so which is equal to 4.
01:07
Next dou f by that is the function is nothing but z equals y square plus 4x.
01:16
So, here dou f by dou y which is equal to 2y.
01:22
On substituting in this then we get s equals double integral over r square root of 1 plus 4 the whole square plus 2y the whole square into dx dy which is equal to double integral over r square root of 17 plus 4y square dx dy...