# Why isn't the answer of this surface integral $24$?

Evaluate the surface integral

$$\int_{S} (z + x^{2}y)\,dS.$$

$S$ is the part of the cylinder $y^{2} + z^{2} = 4$ that lies between the planes $x = 0$ and $x = 3$ in the first octant.

I did the calculation

Integrate[(Sin[v] + u^2 Cos[v]) 2, {v, 0, Pi/2}, {u, 0, 3}]

and got the answer $24$.

But it showed it is not correct, why?

• You should explain how you got from the integral in the problem to the calculation you did. It would be much easier if define how your $u,v$ coordinates relate to $x,y,z$. It looks like you are using cylindrical coordinates with $u$ along $x$ and $v$ the angle. Nowhere do you use the fact that you are integrating over the cylindrical surface, so your answer cannot be right. If you write it up, you might see where the problem is and certainly it will be easier for us to see. -1 – Ross Millikan Apr 11 '17 at 3:23

The given region can be parameterized by

$$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\langle u,2\cos v,2\sin v\rangle$$

where $0\le u\le3$ and $0\le v\le\dfrac\pi2$, as you suggest in your integration setup. However, just replacing $x,y,z$ with their parametric forms isn't enough. You also have to account for the change in coordinates, similar to the way you need to compute the Jacobian in an area/volume integral upon changing coordinates.

This is accounted for by taking the norm of the cross product of the partial derivatives of $\mathbf r(u,v)$ with respect to its independent variables.

$$\mathbf r_u=\langle 1,0,0\rangle$$ $$\mathbf r_v=\langle 0,-2\sin v,2\cos v\rangle$$ $$\implies\left\|\mathbf r_u\times\mathbf r_v\right\|=2\sqrt{\cos^2v+\sin^2v}=2$$

So the integral is given by

$$\iint_S(z+x^2y)\,\mathrm dS=\color{red}2\int_{u=0}^{u=3}\int_{v=0}^{v=\pi/2}(2\sin v+2u^2\cos v)\,\mathrm dv\,\mathrm du=\color{red}{48}$$

so that, in fact, the value of the integral should be twice what you found.