5

To find:

$$I =\int\int_Rx(1+y^2)^{\frac{-1}{2}}dA$$

R is the region in the first quadrant enclosed by $y=x^2$, $y=4$, and $x=0$

$$y=x^2, y=4,x=0, (x= y^\frac{1}{2})$$

$$R=((x,y), 0 \le y \le 4, x^2 \le y \le 4)$$

i.e.

$$R=((x,y), 0 \le x \le 2, x^2 \le y \le 4)$$

$$I=\int_0^2\int_{x^2}^4x(1+y^2)^{\frac{-1}{2}}\,dy\,dx$$

Where to from here?

27182818sheena
  • 315

3 Answers3

3

It can be done very simply like this \begin{align} I=\int_{0}^{4}\!\int_{0}^{\sqrt {y}}\!{\frac {x}{\sqrt {{y}^{2}+1}}} \,{\rm d}x\,{\rm d}y \end{align} \begin{align} =\int_{0}^{4}\!{\frac {\int_{0}^{\sqrt {y}}\!x\,{\rm d}x}{\sqrt {{y}^{2 }+1}}}\,{\rm d}y \end{align} \begin{align} =\int_{0}^{4}\!1/2\,{\frac {y}{\sqrt {{y}^{2}+1}}}\,{\rm d}y \end{align} using $[y^2+1 = u^2]$ \begin{align} =1/2\,\int_{1}^{\sqrt {17}}\!1\,{\rm d}u \end{align} \begin{align} \approx 1.561552813 \end{align}

zhk
  • 585
  • 7
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  • You can use double dollar signs instead of single dollar signs to get displayed equations, and sections contained in \begin{align} and \end{align} to align the equality signs in consecutive equations. – joriki May 06 '16 at 08:07
  • The last equal sign should be an approximation sign. – joriki May 06 '16 at 08:08
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\begin{eqnarray*} % \nonumber to remove numbering (before each equation) I&=& \int_0^4\left(\int_0^{\sqrt{y}}\frac{x}{\sqrt{1+y^2}}dx\right)dy\\ &=& \int_0^4\frac{1}{\sqrt{1+y^2}}\left(\int_0^{\sqrt{y}}xdx\right)dy \\ &=& \frac{1}{2} \int_0^4\frac{y}{\sqrt{1+y^2}} dy\\ &=&\frac{1}{2}\left[\sqrt{1+y^2}\right]^4_0\\ &=&\frac{1}{2}\left(\sqrt{17}-1\right) \end{eqnarray*}

Driss Alami Louati
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As you defined well, $$ R=\left\{ (x,y) \in \mathbb{R}^2: 0\leq x \leq 2, x^2 \leq y \leq 4 \right\}. $$ So $$ I = \int_{R} \frac{x}{\sqrt{1+y^2}} \mathrm{d}A = \int_0^2 \left( \int_{x^2}^4 \frac{x}{\sqrt{1+y^2}} \mathrm{d}y \right) \mathrm{d}x $$ Here, either you do a change of variables or you check in any table that: $$ \frac{\mathrm{d}}{\mathrm{d}z}\sinh^{-1}(z)=\frac{1}{\sqrt{1+y^2}} $$ So \begin{align} I &= \int_0^2 \left[ x\sinh^{-1}(y) \right] ^{y=4}_{y=x^2} \mathrm{d}x \\ &= \int_0^2 \left( x\sinh^{-1}(4) -x\sinh^{-1}(x^2)\right) \mathrm{d}x = \\ &= \left[\frac{x^2}{2}\sinh^{-1}(4)\right]^{x=0}_{x=2} - \int_0^2x\sinh^{-1}(x^2) \mathrm{d}x \\ &= 2\sinh^{-1}(4) - \frac{1}{2}\left[x^2 \sinh^{-1} \left(x^2\right) -\sqrt{x^4+1}\right]^{x=0}_{x=2}\\ &= 2\sinh^{-1}(4) - \frac{1}{2} \left(4 \sinh ^{-1}(4)+1-\sqrt{17}\right) \\ &= \frac{\sqrt{17}-1}{2}. \end{align}

Guilherme Thompson
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