How do I integrate the first part of this double integral?

[Lorentz_F]

Why hello,

I have come across the surface integral \int\int(\frac{xz}{\sqrt{16-x^{2}}+x})dzdx

My query is, which type of integration do I use to solve the first part of this double integral. The solution is: \int(4x+8)dz = 90

 

[HallsofIvy]

Why hello,

I have come across the surface integral \int\int(\frac{xz}{\sqrt{16-x^{2}}+x})dzdx

My query is, which type of integration do I use to solve the first part of this double integral. The solution is: \int(4x+8)dz = 90

No one can tell you why the answer is that particular number if you don't tell us what the limits of integration are! And, while it is easy to integrate with respect to z, it the limits of integration on z are functions of x, that would strongly affect how you did the integral with respect to x. Please tell us what the limits of integration are.

 

[Lorentz_F]

well the limits are Z=0, z=5, x=0, x=4, but what I'm after is what type of integration to use i.e. by substitution, parts, product malarchy or what? Just from looking at the bit \left(\frac{xz}{\sqrt{16-x^{2}}}+x\right) as I havn't the foggiest.

Thank you.

 

[Lorentz_F]

oh and the fist equation is wrong, the bit in the middle is as I've just posted, sorry, and i see your point on the function of x limits now

 

[HallsofIvy]

Now I'm wondering what "product malarchy" means! It sounds dangerous.

\int_{x= 0}^4\int_{x= 0}^5 \left(\frac{xz}{\sqrt{16- x^2}}+ x\right)dzdx

The simplest thing to do is to separate the integrals as:
\left(\int_{x= 0}^4 \frac{x}{\sqrt{16- x^2}} dx\right)\left(\int_{z=0}^5 zdz\right)+ \left(\int_{x=0}^4 x dx\right)\left(\int_{z=0}^5 dz\right)

To integrate the first of those, let u= 16- x^2. The others should be easy.