00:01
Hello everyone.
00:02
We are given a portion of a cone z to be root of x square plus y square divided by 3 between z is equal to 1 and z is equal to 4 divided by 3.
00:14
We need to find the parameterization of the above portion of the cone and using which we need to find the integral.
00:21
So we see that we have to find the surface as a double integral.
00:26
Therefore, so let's push through the solution.
00:30
First of all, we put x is equal to r cosine theta and we have y will be equal to r sine theta.
00:52
So now we see that theta varies between 0 and 2 pi.
01:00
So we have, after converting this, so how do we convert the given curve? we have to put in place of x r cosine theta and in place of y, r sine theta.
01:15
So a z will become root of r square, since x square, i have to write r square, cosine square theta, plus r square sine square theta root divided by three so i see that r square is common so i can take it out and there's square root so i will get only r and we know that will be equal to one so that will become root one which is going to give me one divided by three so now we have z value z is equal to are divided by 3.
02:04
So now we have to find the limit points, right? so we are given that z is lying between z is equal to 1 and z is equal to 4 divided by 3.
02:17
So we can write that given that z is equal to 1 and z is equal to 4.
02:32
The curve is going to the curve is going to be between 1 divided by 3.
02:42
So now after simplifying, we have to convert to the polar, that is parametizing.
02:49
So we will have 1 less than or equal to z is nothing but r divided by 3, which is less than or equal to 4 divided by 3.
02:58
On simplification, we will have 3 less than or equal to r less than or equal to.
03:07
So now let's write the area of the surface integral...