Question

Double integral of f(x,y)=10x^4 y using the figure attached

Answer 1

using the figure eh...

I take it we need to calculate the bounds of x and y then

Answer 2

x is from x=y to 4; and y is from 0 to 2

Answer 3

\[\int_{0}^{2}\int_{y}^{4}\ f(x)\ dx.dy\]

Answer 4

if we flip it around we can get:

y is from 0 to x; and x is from 0 to 4

Answer 5

\[\int_{0}^{4}\int_{0}^{x}\ f(y) \ dy.dx\]

Answer 6

is that Fubini's Theorem ?? I think someone mentioned that the other day

Answer 7

i got know idea whose thrm it is :)

Answer 8

haha okay :) nm

Answer 9

but let me see what I can do to it... i need the practice :)

\[\int_{0}^{2}\int_{y}^{4}(10x^4 y).dxdy\]
\[\int_{0}^{2}(2(4)^5 y)-(2y^5y)dy\]
\[\int_{0}^{2}(2048y)dy-\int_{0}^{2}(2y^6)dy\]

Answer 10

\[1024(2)^2-1024(0)^2-[\frac{2(2)^7}{7}-\frac{2(0)^7}{7}]\]

Answer 11

with any luck; that makes sense :)

Answer 12

yes I follow, I was doing it on paper :)

Answer 13

the thing that threw me off the most was trying to get the bounds from the graph, prob the simplest part haha.

Answer 14

how do I tell the min and max from the double intergral ?

Answer 15

min and max from integration? cant say that I recall; or are you refering to the surface?

Answer 16

bounds i calculate like this

Answer 17

oh whoops sorry, I was getting my quesiton confused.

Answer 18

the answer was right! thanks again that made it much clearer. I had wrong bounds :)

Answer 19

yay!!

Answer 20

try the bounds both ways and see if you get the same results; you should after all

Answer 21

okay, I need the practice!

Answer 22

\[\int_{0}^{4}\int_{0}^{x}(10x^4y)\ dy.dx\]
\[\int_{0}^{4}(5x^4x^2-5x^40^2)\ dx\]
\[\int_{0}^{4}(5x^6)\ dx\]
\[\frac{5(4)^7}{7}-\frac{5(0)^7}{7}\]
if i did it right

Answer 23

perhaps the bounds there are 0 to 2 for the y?

Answer 24

well, at least one way worked ;)