Question

come someone show me the steps! double integration problem. problem and solution found here: https://www.wolframalpha.com/input/?i=integrate%201/%28sqrt%28%28y^2%29%2B16%29%29%20dx%20dy%20from%20y=0%20to%203%20from%20x=-abs%28y%29%20to%20abs%28y%29

Answer 1

first step is just integrate wrt x. since there is no x in the integrand, you're just integrating dx.

Answer 2

yes, I know. I am not seeing how to deal with the steps beyond that.

Answer 3

next use u substitution u=y^2+16, du=2y dy

Answer 4

using the fact that abs(y)=y for the given integration limits

Answer 5

the inner integral is 2y/sqrt(y^2+16)

Answer 6

using that information I end up with integral of 0 to 25 of 1/sqrt(u) and end with the answer 10... the answer is 2 though... what am I doing wrong?

Answer 7

the limits are 0 to 3 on y

\[ \int_0^3 u^{-\frac{1}{2}} du \]

Answer 8

I change the limits of integration when using u substitution

Answer 9

ok, then the lower limit is 4 ?

Answer 10

why 4? don't you plug in the limits to what u is equal to... y^2+16 which would actually end with 16 to 25

Answer 11

I don't know, no matter what with u substitution I don't seem to be getting the answer of 2

Answer 12

\[\int_0^3\int_{-|y|}^{|y|}\frac{1}{\sqrt{y^2+16}}~dx~dy\\
\int_0^3\frac{1}{\sqrt{y^2+16}}\left(|y|-(-|y|)\right)~dy\]
Since \(y>0\), \(|y|=y\):
\[\int_0^3\frac{2y}{\sqrt{y^2+16}}~dy\]
Let \(u=y^2+16\), then \(\dfrac{1}{2}du=y~dy\):
\[\frac{1}{2}\int_{16}^{25}\frac{2}{\sqrt{u}}~du\\
\int_{16}^{25}u^{-1/2}~du\]

Answer 13

yes, the limits are 16 to 25... I was thinking u= sqr(stuff) \[ \int_{16}^{25} u^{-\frac{1}{2}} du = 2 \sqrt{u} | = 2(5-4) = 2 \]

Answer 14

thank you both so much!!!!