# Thread: Double integral set up #2

1. ## Double integral set up #2

so I am given

$\displaystyle \int_{0}^{2}\int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2} \Bigg(\frac{5}{\sqrt{2+x}} + y^3} \Bigg)dydx$

Could I do this as

$\displaystyle 2\int_{0}^{2}\int_{0}^{3\sqrt{4-x^2}/2} \Bigg(\frac{5}{\sqrt{2+x}} + y^3} \Bigg)dydx$

I remember doing some integrals in calculus 2 where you could do that. I think they were stuff dealing with the polar coordinates though

2. ## Re: Double integral set up #2

Originally Posted by Jonroberts74
so I am given

$\displaystyle \int_{0}^{2}\int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2} \Bigg(\frac{5}{\sqrt{2+x}} + y^3} \Bigg)dydx$

Could I do this as

$\displaystyle 2\int_{0}^{2}\int_{0}^{3\sqrt{4-x^2}/2} \Bigg(\frac{5}{\sqrt{2+x}} + y^3} \Bigg)dydx$

I remember doing some integrals in calculus 2 where you could do that. I think they were stuff dealing with the polar coordinates though
You need to do more than just guess.

$\displaystyle{\int_0^2 \int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2}}\left(\dfrac{5}{\sqrt{2+x}}+y^3\right)~dy~dx=$

$\displaystyle{\int_0^2}\dfrac{5}{\sqrt{2+x}} \left( \displaystyle{\int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2}}y^3~dy\right) ~dx$

Now the integral over $dy$ is that of an odd function over limits symmetric about zero.

What does that tell you about it's value?

3. ## Re: Double integral set up #2

it goes to zero when an odd function is integrated over the symmetric about zero

4. ## Re: Double integral set up #2

Correct, so what is the value of the double integral?

5. ## Re: Double integral set up #2

after integrating for y

$\displaystyle \int_{0}^{2} 15\sqrt{2-x} dx = -10(2-x)^{3/2} \Bigg|_{0}^{2} = 20\sqrt{2}$

6. ## Re: Double integral set up #2

Originally Posted by Jonroberts74
after integrating for y

$\displaystyle \int_{0}^{2} 15\sqrt{2-x} dx = -10(2-x)^{3/2} \Bigg|_{0}^{2} = 20\sqrt{2}$
correct.