What is the area of the part of the surface $z=yx$ bounded by $x^2+y^2=1$? A parametrization of the part of the surface $z=yx$ bounded by $x^2+y^2=1$ is
\begin{align}
x &amp;= u \cos v \\
y &amp;= u \sin v \\
z &amp;= \frac12 u^2 \sin 2v,
\end{align}
or
$$r(u,v)=u \cos v \, {\bf i} + u \sin v \, {\bf j} + \frac12 u^2 \sin 2v \, {\bf k}, \quad 0&lt;v&lt;2\pi, 0&lt;u&lt;1.$$

The norm of the cross product of $r_u$ and $r_v$ is $$\sqrt {u^4+u^2} = u\sqrt {u^2+1},$$
so
$$\int^{2\pi}_0\int_0^1u\sqrt {u^2+1}dudv=\int^{2\pi}u\sqrt {u^2+1}=\frac13((4\pi^2+1)^\frac23-1)$$

However, the answer showed in my text book is $\frac{2\pi}{3}(2\sqrt 2-1)$, so I don't know whether I did it wrong or there is a mistake in my textbook.
The first integral equality is not correct---you seem to be integrating with respect to $dv$ but over the interval for $u$.

Since the integrand does not depend on $v$, we can write
$$\int_0^{2 \pi} \int_0^1 u \sqrt{u^2 + 1} du\,dv = \int_0^{2 \pi} dv \int_0^1 u \sqrt{u^2 + 1} du = 2 \pi \int_0^1 u \sqrt{u^2 + 1} du$$
$I=\int \int_Vxy\ dx\ dy = \int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^\sqrt{1-x^2} xy\ dy\ dx$

Change of coordinates -

$x=r\cos \theta$

$y=r\sin \theta$

Jacobian -

$I=\int \int r^2\cos\theta \sin\theta\ \left\|d\frac{x,y}{r,\theta}\right\|\ d\theta\ dr$

$\left\|d\frac{x,y}{r,\theta}\right\|=\left|\begin{array}{c}\cos \theta&amp;\sin\theta\\ -r\sin \theta&amp;r\cos \theta\end{array}\right|=r$

$I=\int_0^1 \int_0^{2\pi} r^3\cos\theta \sin\theta\ d\theta\ dr=\int r^3\cdot0\ dr=0$

In retrospect, it is easy to see this is zero geometrically, since $xy+(-x)y=0$.