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Calculate surface area of $z = x^2, \quad 0 \leq y \leq x \leq 1$

My attempt:

I parametrize the surface with $$r(x,y) = (x,y,x^2),$$ where the domain E is given by: $$0 \leq x\leq1,$$ $$0 \leq y \leq x.$$ I find $|r'_x \times r'_y|$: $$|r'_x \times r'_y|$$ $$|(1,0,2x) \times (0,1,0)|= |(-2x,0,1)| = \sqrt{4x^2+1}$$

The area is given by $$\iint_E(\sqrt{4x^2+1})dxdy,$$ but I can't compute this integral and it doesn't seem to be computable with reasonable complexity according to wolfram alpha.

What am I doing wrong? The answer is supposed to be: $\frac{5\sqrt{5}-1}{12}$

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    $\begingroup$ $\int\int_E \sqrt{4x^2 +1} dx dy = \int_0^1 \int_y^1 \sqrt{4x^2 + 1} dx dy = \int_0^1 \int_0^x \sqrt{4x^2 +1} dy dx = \int_0^1 x \sqrt{4x^2 + 1}dx$ $\endgroup$
    – Cauchy
    Aug 9, 2017 at 14:57
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    $\begingroup$ @Cauchy Thanks, that solved it. Well this was an unnecessary question. $\endgroup$
    – Heuristics
    Aug 9, 2017 at 15:07

1 Answer 1

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$$S={\iint_{E}{\sqrt{1+4x^2}}dxdy}$$ $$S={\int_{x=0}^1\int_{y=0}^{x}\sqrt{1+4x^2}dydx}$$ $$S={\int_{x=0}^{1}\sqrt{1+4x^2}dx}{\int_{y=0}^xdy}$$ $$S={\int_{x=0}^{1}{x\sqrt{1+4x^2}}dx}$$ By substitution $1+4x^2=u^2$$\implies$ $4xdx=udu$ $$S={\int_{u=1}^{\sqrt5}{u^2\over4}du}$$ $${S} = {(5\sqrt5-1)\over{12}}$$

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