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[dusty ermine]

is this correct?

[covert coral]

,align\text{Volume}&=\int_1^4\pi y^2\dd{x}
\&=\int_1^4\pi\sqrt{x}^2\dd{x}
\&=\int_1^4x\dd{x};\pi
\&=\qty(\frac{4^2}2-\frac{1^2}2)\pi
\&=\boxed{\frac{15}2\pi}
\\text{Surface area}&=\int_1^42\pi y\sqrt{1+\qty(\dv{y}{x})^2}\dd{x}
\&=\int_1^42\pi\sqrt x\sqrt{1+\qty(\frac1{2\sqrt x})^2}\dd{x}
\&=2\int_1^4\sqrt x\sqrt{1+\frac1{4x}}\dd{x};\pi
\&=2\int_1^4\sqrt{x+\frac14}\dd{x};\pi
\&=2\qty(\frac23\qty(4+\frac14)^{3/2}-\frac23\qty(1+\frac14)^{3/2})\pi
\&=\frac43\qty(\frac{17^{3/2}}8-\frac{5^{3/2}}8)\pi
\&=\boxed{\frac{17^{3/2}-5^{3/2}}6\pi}

[vestal cloud]

mtt07734

[covert coral]

.solved deleted user