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Is my procedure correct?

Calculate paraboloid area portion of equations

$$P \equiv (u \cos v, u \sin v, u^2) $$

with $0 \leq v \leq \dfrac{\pi}{4}$ and $0 \leq u \leq \dfrac{1}{2}\tan v$

This is my attempt:

$\bullet$ vector first derivatives:

$$P'_u = (\cos v, \sin v, 2u)$$

$$P'_v = (-u\sin v, u\cos v, 0)$$

$\bullet$ cross product between $P'_u$ and $P'_v$

$$P'_u \wedge P'_v = (-2u^2\cos v, 2u^2\sin v, u)$$

$\bullet$ cross product module

$$N(u,v) = u\sqrt{4u^2 + 1} $$

$\bullet$ double integral

$$\displaystyle \iint_P u\sqrt{4u^2 + 1} \; du \; dv = \int^{\beta(v)}_{\alpha(v)} u\sqrt{4u^2 + 1} \; du$$

where $\alpha(v) = 0$ and $\beta(v) = \dfrac{1}{2} \tan \dfrac{\pi}{4} = \dfrac{1}{2}$

$$\displaystyle \int^{\beta(v)}_{\alpha(v)} u\sqrt{4u^2 + 1} \; du = \displaystyle \dfrac{1}{12} (4u^2 + 1)^{3/2} \Big|^{u=1/2}_{u=0} = \dfrac{1}{12}(2\sqrt{2} - 1)$$

Thank you in advance

Community
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user3204810
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1 Answers1

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The partial derivative $P_u'$ shouldn't have a minus sign in the second component. Otherwise, everything was fine until you inserted everything into the integral. Notice that your bound for $u$ depends on $v$, so you should have \begin{align} \int_P \, \mathrm dA &= \int_{v=0}^{\pi/4} \int_{u=0}^{(\tan v)/2} u\sqrt{4u^2+1} \,\mathrm du \, \mathrm dv\\ &= \frac{1}{12}\int_{v=0}^{\pi/4}\left((\tan^2v+1)^{3/2}- 1\right) \, \mathrm dv \\ &= \frac{1}{12}\int_{v=0}^{\pi/4} \sec^3 v \, \mathrm dv - \frac{\pi}{48}, \end{align} etc.

MSDG
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