Calculating antenna directivity

[JamesGoh]

Homework Statement

An antenna has a normalised E-filed pattern, En where \theta = vertical angle as measured from z-axis and \phi = azimuth angle measured from x-axis.
Calculate the exact directivity

En has a non-zero value whenever 0 <= \theta <= \pi and 0 <= \phi <= \pi. Elsewhere, En is zero


The correct answer is 6 , but I cannot get this number

Homework Equations

En = sin(\theta)sin(\phi)

Direcitivity is calculating using

D = 4\pi/ ( \int\intPn(\theta,\phi )d\phid\theta )

The Attempt at a Solution

Ok, first of all Pn = En^{2}

Therefore Pn = sin(\theta)^{2}sin(\phi)^{2}

Now sin^{2}(\theta) = 0.5(1 - cos(2\theta)) with respect to \theta

If we perform the integration of the sin^{2}(\theta) terms we should get

\theta/2 - sin(2\theta)/4

Applying the 0 <= \theta <= \pi limits, I got \pi/2 for the first integral. If we integrate the sin(\phi) term, we should get \pi/2 as well, making \pi^{2}/4 the actual answer for

\int\intPn(\theta,\phi )d\phid\theta )

Substituting \pi^{2}/4 into the denomiator part of D, I get 16/\pi which obviously is not right. I am not sure where I could be going wrong

 

[JamesGoh]

If we perform the integration of the sin^{2}(\theta) terms we should get

\theta/2 - sin(2\theta)/4

Applying the 0 <= \theta <= \pi limits, I got \pi/2 for the first integral. If we integrate the sin(\phi) term, we should get \pi/2 as well, making \pi^{2}/4 the actual answer for

\int\intPn(\theta,\phi )d\phid\theta

 

[JamesGoh]

Guys, please refer to my 2nd most outlining the integration process

cheers and thanks in advance

 

[JamesGoh]

Homework Statement

An antenna has a normalised E-filed pattern, En where \theta = vertical angle as measured from z-axis and \phi = azimuth angle measured from x-axis.
Calculate the exact directivity

En has a non-zero value whenever 0 &lt;= \theta &lt;= \pi and 0 &lt;= \phi &lt;= \pi. Elsewhere, En is zero


The correct answer is 6 , but I cannot get this number

Homework Equations

En = sin(\theta)sin(\phi)

Direcitivity is calculating using

D = 4\pi/ \int\intPn(\theta,\phi )d\phid\theta

The Attempt at a Solution

Ok, first of all Pn = En^{2}

Therefore Pn = sin(\theta)^{2}sin(\phi)^{2}

Now sin^{2}(\theta) = 0.5(1 - cos(2\theta)) with respect to \theta

If we perform the integration of the sin^{2}(\theta) terms we should get

\theta/2 - sin(2\theta)/4

Applying the 0 &lt;= \theta &lt;= \pi limits, I got \pi/2 for the first integral. If we integrate the sin(\phi) term, we should get \pi/2 as well, making \pi^{2}/4 the actual answer for

\int\intPn(\theta,\phi )d\phid\theta

Substituting \pi^{2}/4 into the denomiator part of D, I get 16/\pi which obviously is not right. I am not sure where I could be going wrong