## Calculus: Early Transcendentals 8th Edition

$\dfrac{21 \sqrt {21} -17 \sqrt {17}}{12}$
Here, we have $\iint_S x dS =\iint_{R} x \times \sqrt {17 +4x^2} dA$ $=\iint_{R} x \times \sqrt {17 +4x^2} dx dz$ $=\int_{0}^{1} dz \int_0^1 \sqrt {17 +4x^2} dx$ $=\int_{0}^{1} x \sqrt {17 +4x^2} dx$ Plug $a=17+4x^2 \implies da=8r dr$ $= \int_{17}^{21} \int_1^{17} a^{1/2} \times \dfrac{da}{8}$ $=\dfrac{1}{8} [(2/3) a^{3/2} ]_{17}^{21}$ $=\dfrac{1}{12} [a \sqrt a]_{17}^{21}$ $=(1/12) (21 \sqrt {21} -17 \sqrt {17})$ $=\dfrac{21 \sqrt {21} -17 \sqrt {17}}{12}$