# double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by

• May 6th 2014, 10:42 PM
SNAKE
double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by
bounded by y=x^2, y=4, x=0. picture: graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha

i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx

then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2

but second integral, how to integrate x*arcsinh(x^2).

edit: is this how
[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2
• May 6th 2014, 11:25 PM
Idea
Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by
Quote:

Originally Posted by SNAKE
bounded by y=x^2, y=4, x=0. picture: graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha

i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx

then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2

but second integral, how to integrate x*arcsinh(x^2).

edit: is this how
[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2

reverse the order of integration?
• May 7th 2014, 12:00 AM
Prove It
Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by
Quote:

Originally Posted by SNAKE
bounded by y=x^2, y=4, x=0. picture: graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha

i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx

then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2

but second integral, how to integrate x*arcsinh(x^2).

edit: is this how
[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2

Horizontal strips are a better choice than vertical in this case. So \displaystyle \begin{align*} 0 \leq x \leq \sqrt{y} \end{align*} and \displaystyle \begin{align*} 0 \leq y \leq 4 \end{align*}, so that means your double integral is

\displaystyle \begin{align*} \int_0^4{\int_0^{\sqrt{y}}{x \left( 1 + y^2 \right) ^{-\frac{1}{2}} \,dx}\,dy} &= \int_0^4{ \left[ \frac{x^2}{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right]_0^{\sqrt{y}} \,dy } \\ &= \int_0^4{ \left[ \frac{ \left( \sqrt{y} \right) ^2 }{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right] - \left[ \frac{0^2}{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right] \,dy } \\ &= \frac{1}{2}\int_0^4{ y \, \left( 1 + y^2 \right) ^{-\frac{1}{2}} \, dy }\\ &= \frac{1}{4} \int_0^4{ 2y \, \left( 1 + y^2 \right) ^{-\frac{1}{2}} \, dy } \\ &= \frac{1}{4} \int_1^{17}{ u^{-\frac{1}{2}} \, du } \textrm{ when we substitute } u =1 + y^2 \\ &= \frac{1}{4} \left[ 2u^{\frac{1}{2}} \right]_1^{17} \\ &= \frac{1}{4} \left[ 2 \left( 17 \right) ^{\frac{1}{2}} - 2 \left( 1 \right) ^{\frac{1}{2}} \right] \\ &= \frac{1}{2} \left( \sqrt{17} - 1 \right) \end{align*}
• May 7th 2014, 12:05 AM
SNAKE
Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by
ahh, i considered doing the horizontal method but I didn't see how it would alleviate the difficult integral but seeing it written out like this, you have Proven It!!

I am adding extra smilies to my sig to show my appreciation :)
• May 7th 2014, 07:20 AM
HallsofIvy
Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by
Quote:

Originally Posted by SNAKE
bounded by y=x^2, y=4, x=0. picture: graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha

i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx

then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2

but second integral, how to integrate x*arcsinh(x^2).

edit: is this how
[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2

Reversing the order of integration, as both Idea and Prove It said is the best way to handle this integral. But to answer you last question, "how to integrate x arcsinh(x^2)" let u= x^2 so [tex]\int x arcsinh(x) dx[tex] becomes $\frac{1}{2}\int arcsinh(u)du$ and that can be looked up in a table of integrals.
(Lists of integrals - Wikipedia, the free encyclopedia)
• May 7th 2014, 06:31 PM
Prove It
Re: double integral of x(1+y^2)^-0.5 over region in first quadrant bounded by
Quote:

Originally Posted by HallsofIvy
Reversing the order of integration, as both Idea and Prove It said is the best way to handle this integral. But to answer you last question, "how to integrate x arcsinh(x^2)" let u= x^2 so [tex]\int x arcsinh(x) dx[tex] becomes $\frac{1}{2}\int arcsinh(u)du$ and that can be looked up in a table of integrals.
(Lists of integrals - Wikipedia, the free encyclopedia)

Or otherwise derived (note I've left out the integration constants until the end)...

Let's consider \displaystyle \begin{align*} \int{ \frac{2x}{\sqrt{1 + x^2}} \, dx} = \int{ 2x \left( 1 + x^2 \right) ^{-\frac{1}{2}} \,dx} \end{align*}. With the substitution \displaystyle \begin{align*} u = 1 + x^2 \implies du = 2x\,dx \end{align*} we can see \displaystyle \begin{align*} \int{ 2x \left( 1 + x^2 \right) ^{-\frac{1}{2}} \, dx} = \int{ u^{-\frac{1}{2}} \, du } = 2u^{\frac{1}{2}} = 2 \, \sqrt{ 1 + x^2 } \end{align*}.

We might also consider integration by parts:

\displaystyle \begin{align*} \int{ \frac{2x}{\sqrt{1 + x^2}} \, dx} &= \int{ 2x \, \frac{1}{\sqrt{1 + x^2}} \, dx} \\ &= 2x\,\textrm{arsinh}\,{(x)} - \int{ 2\,\textrm{arsinh}\,{(x)}\,dx } \\ &= 2x\,\textrm{arsinh}\,{(x)} - 2\int{ \textrm{arsinh}\,{(x)}\,dx } \end{align*}

and thus

\displaystyle \begin{align*} 2x\,\textrm{arsinh}\,{(x)} - 2\int{ \textrm{arsinh}\,{(x)}\,dx} &= 2\,\sqrt{1 + x^2} \\ x\,\textrm{arsinh}\,{(x)} - \int{ \textrm{arsinh}\,{(x)}\,dx} &= \sqrt{1 + x^2} \\ \int{\textrm{arsinh}\,{(x)}\,dx} &= x\,\textrm{arsinh}\,{(x)} - \sqrt{1 + x^2} + C \end{align*}