15.3.50 Double integral of circle and graph

[karush]

$\displaystyle
\int_{0}^{1}
\int_{0}^{\sqrt{1-x^2}}
\sqrt{x^2+y^2}
\, dydx=\frac{\pi}{6}$

this was the W|A answer
but how ?

also supposed to graph this
but didn't know the input for desmos

 

[I like Serena]

Re: 15.3.50 dbl int of circle and graph

$\displaystyle
\int_{0}^{1}
\int_{0}^{\sqrt{1-x^2}}
\sqrt{x^2+y^2}
\, dydx=\frac{\pi}{6}$

this was the W|A answer
but how ?

also supposed to graph this
but didn't know the input for desmos

Hi karush!

The occurrence of $\sqrt{x^2+y^2}$ strongly suggests that polar coordinates are in order.
How about substituting $x=r\cos\phi, y=r\sin\phi$?

As for graphing, I guess we would want to graph the boundaries of the area that we integrate.
Can we tell which function and line segments describe those boundaries, so that we can graph them in Desmos?

 

[karush]

Re: 15.3.50 dbl int of circle and graph

$\displaystyle
\int_{0}^{1}
\int_{0}^{r\sin\theta}
\sqrt{r\cos^2\theta+r\sin^2 \theta}
\, d\theta dr$
are you sugesting this