00:01
Hello students today we will discuss about this question.
00:04
In this question we are given that the integration minus 3 to 0 2 ,000, 2nd of 9 minus x square, integration 9 minus x square, integration 0 to 3 x, x, y square d z, d y, d x, we need to evaluate these integral by changing the cylindrical coordinates.
00:29
So now we know that the cylindrical coordinates that are x that is equals to r cost theta, y that is equals to r sine theta, and z that is equals to z, and dg, d, y, d, d, x, that is equal to d, d, x, d, d, x, d, d, x, d, d, d, tater.
00:50
And we know that x square plus y square that is equals to r square.
00:55
So let i that is equal to integration of my, 2 to 0 integration minus square root of 9 minus x square to plus 9 minus square integration 0 to 3 x y square d z, d y d x, d x, now x that is equal to minus 3 to x that is equals to 0, y that is equal to minus square root of 9 minus x squared 2 plus 9 minus x squared and z that is equals to 0 to z that is equal to 3 so therefore an x square plus y square that is equals to 9 so therefore zero is less than r is less than 3 and therefore pi divide by 2 is less than or equals to tita is less than or equals to 3 pi divided by 2 so therefore i that is equals to integration of pi divide by 2 to 3 pi divide by 2 integration 0 to 3 integration 0 to 3 r cost theta r square sine square theta multiplied by r d z d d theta so that is equals to integration pi divide by 2 2 2 5 divide by 2 cost theta sine square tita d theta integration 0 to 3 r raise to 4 z limit will be 0 to 3 dr.
02:33
So therefore we will get let sign theta that is equals to t and cost theta d theta that is equals to d t.
02:44
So therefore theta that is equals to pi divide by 2 that gives 1 and theta that is equals to 3 pi divide by 2 that gives minus 1...