- anonymous

int(int(6xy^3 x=y..1) y=0..1) dxdy

- jamiebookeater

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- anonymous

Is your problem\[\int\limits_{0}^{1}\int\limits_{y}^{1}6xy^3dxdy\]?

- anonymous

yes it is

- anonymous

Okay, just give me a sec to do something non-mathematical. I can help you.

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## More answers

- anonymous

ok, cool. i got 1/4, can you tell me if that's right? when ever you're done though

- anonymous

Okay...I scratched it out and got 1/4 too. You know what you're doing.

- anonymous

yeah but i'm have a problem with another one...is it ok to ask you one more?

- anonymous

sure

- anonymous

ok it's another double...int(int(5sin(x+y) x=0..pi/2) y=0..pi/2) dxdy

- anonymous

I got 10. Is that what you got?

- anonymous

i keep getting stuck...this is the third time trying to do it again

- anonymous

It might be easier for me to scan what I did and attach then write it out...have a look through if and ask questions.

- anonymous

perfect thanks

- anonymous

Everything under the red line.

##### 1 Attachment

- anonymous

sorry im comparing right now...

- anonymous

That's okay. I'll be online for a while. Just post when you're ready. If I don't respond 'immediately', I'm away from the computer.

- anonymous

ok i see it

- anonymous

I think you're getting trapped by not just accepting that the other variable is just a constant when you integrate over the other in a double integral.

- anonymous

your fourth step is different from mine. i dont understand how you went from cos (pi/2 + y) to it becoming sin y

- anonymous

We get very used to thinking of x and y as things that vary, rather than stand still.

- anonymous

You can use the double angle formula. You'll see I did that expansion in the top right corner (under the red line).

- anonymous

\[\cos(\frac{\pi}{2}+y)=\cos \frac{\pi}{2}\cos y - \sin \frac{\pi}{2}\sin y\]\[=0 \times \cos y + (- 1) \times \sin y=-\sin y\]

- anonymous

thank you so much. i would have never remembered that rule

- anonymous

no probs.

- anonymous

when you're done and you're satisfied, it'd be great if you could click the 'become a fan' link :)

- anonymous

just did it

- anonymous

cheers

- anonymous

so you're getting the same answer?

- anonymous

yes, after remembering the rule it all made sense

- anonymous

good.

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