- #1

- 6

- 0

double integral (e^(y^3)) dy dx

Where dy is evaluated from sqrt(x/3) to 1

.....and dx is evaluated from 0 to 3.

I am lost.

I don't even know how to start.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ChargedTaco
- Start date

- #1

- 6

- 0

double integral (e^(y^3)) dy dx

Where dy is evaluated from sqrt(x/3) to 1

.....and dx is evaluated from 0 to 3.

I am lost.

I don't even know how to start.

- #2

mjsd

Homework Helper

- 726

- 3

double integral is just normal integrals in disguise, so when ou are doing the "y"-integral, you treat all "x"s in the integrand as just a constant, and while doing the "x"-integral you treat all "y"s as constant. Then, everything would then be exactly the same as a normal single integral. In many cases, whether you do the x or y integral first does not matter.

in your case i think you should do y first

- #3

- 13,194

- 764

The "y" integration doesn't look pretty. It's basically

[tex] \int_{a}^{b} e^{y^3}{}dy [/tex]

[tex] \int_{a}^{b} e^{y^3}{}dy [/tex]

- #4

- 6

- 0

If I integrate x first then I'll end up with having to

integrate 3e^(y^3)dy....and I again I don't know how to do this.

Thanks.

- #5

- 6

- 0

Here it is.

[tex]\int_{0}^{3} \int_{sqrt(x/3)}^{1} e^{y^3} {}dy} {}dx [/tex]

Yay!! I got the latex code....however this is the only thing I know about this problem.

[tex]\int_{0}^{3} \int_{sqrt(x/3)}^{1} e^{y^3} {}dy} {}dx [/tex]

Yay!! I got the latex code....however this is the only thing I know about this problem.

Last edited:

- #6

- 6

- 0

Nevermind....I got it. Thanks guys.

Share: