int(int(6xy^3 x=y..1) y=0..1) dxdy
Is your problem\[\int\limits_{0}^{1}\int\limits_{y}^{1}6xy^3dxdy\]?
yes it is
Okay, just give me a sec to do something non-mathematical. I can help you.
ok, cool. i got 1/4, can you tell me if that's right? when ever you're done though
Okay...I scratched it out and got 1/4 too. You know what you're doing.
yeah but i'm have a problem with another one...is it ok to ask you one more?
sure
ok it's another double...int(int(5sin(x+y) x=0..pi/2) y=0..pi/2) dxdy
I got 10. Is that what you got?
i keep getting stuck...this is the third time trying to do it again
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.
perfect thanks
Everything under the red line.
sorry im comparing right now...
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.
ok i see it
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.
your fourth step is different from mine. i dont understand how you went from cos (pi/2 + y) to it becoming sin y
We get very used to thinking of x and y as things that vary, rather than stand still.
You can use the double angle formula. You'll see I did that expansion in the top right corner (under the red line).
\[\cos(\frac{\pi}{2}+y)=\cos \frac{\pi}{2}\cos y - \sin \frac{\pi}{2}\sin y\]\[=0 \times \cos y + (- 1) \times \sin y=-\sin y\]
thank you so much. i would have never remembered that rule
no probs.
when you're done and you're satisfied, it'd be great if you could click the 'become a fan' link :)
just did it
cheers
so you're getting the same answer?
yes, after remembering the rule it all made sense
good.
Join our real-time social learning platform and learn together with your friends!