We have, in part (d), the integral in terms of . However the original integral was with respect to so we must answer in terms of . If you have correctly drawn the triangle in question 3(a) you should have in a right-angled triangle with opposite and the hypotenuse equal to . Now use Pythagoras to find the side adjacent to …

When we have done this we can find . These should be in terms of . Now replace the cosines in Q. 3(d) with this expression to get the answer given in part (e).

Regards,

J.P.

If we do as you suggest and let , we have

.

Now note that

using the identity .

Now what about ?

Regards,

J.P.

Having trouble with the project! doing project 4 on areas and volumes. In Q2 im struggling with integrating the u(x). your hint says to substitute x=r sinQ and then use indentities? do you substitute the x= r sin Q so the equation

u(x) is now the sqroot of r sq – rsinQ sq?? ]]>