Taking a wild guess on what the question says...
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
[tex]\displaystyle\iint_Syz\,\mathrm ds[/tex]
where [tex]S[/tex] is the surface parameterized by
[tex]\mathbf r(u,w)=\mathbf r(x(u,w),y(u,w),z(u,w))=\left\langle u^2,u\sin7w,u\cos7w\right\rangle[/tex]
for [tex]0\le u\le1[/tex] and [tex]0\le w\le\dfrac\pi{14}[/tex].
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
The surface integral is equivalent to
[tex]\displaystyle\iint_S y(u,w)z(u,w)\left\|\mathbf r_u\times\mathbf r_w\right\|\,\mathrm du\,\mathrm dw[/tex]
where you have
[tex]\mathbf r_u=\left\langle2u,\sin7w,\cos7w\right\rangle[/tex]
[tex]\mathbf r_w=\left\langle0,7u\cos7w,-7u\sin7w\right\rangle[/tex]
[tex]\implies\mathbf r_u\times\mathbf r_w=\left\langle-7u,14u^2\sin7w,14u^2\cos7w\right\rangle[/tex]
[tex]\implies\left\|\mathbf r_u\times\mathbf r_w\right\|=7\sqrt{u^2+4u^4}[/tex]
The integral then becomes
[tex]\displaystyle7\int_{u=0}^{u=1}\int_{w=0}^{w=\pi/14}u^2\sin7w\cos7w\sqrt{u^2+4u^4}\,\mathrm dw\,\mathrm du[/tex]
[tex]=\displaystyle7\left(\int_{u=0}^{u=1}u^2\sqrt{u^2+4u^4}\,\mathrm du\right)\left(\int_{w=0}^{w=\pi/14}\sin7w\cos7w\,\mathrm dw\right)[/tex]
[tex]=\displaystyle\frac72\left(\int_{u=0}^{u=1}u^2\sqrt{u^2+4u^4}\,\mathrm du\right)\left(\int_{w=0}^{w=\pi/14}\sin14w\,\mathrm dw\right)[/tex]
[tex]=\dfrac{1+25\sqrt5}{240}[/tex]
