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amistre64
 5 years ago
double integral of f(x,y) = 4x^2 y y^2; x,y bounds: y=x^3, and y=cbrt(x) ??
amistre64
 5 years ago
double integral of f(x,y) = 4x^2 y y^2; x,y bounds: y=x^3, and y=cbrt(x) ??

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{a}^{b} \int\limits_{h(x)}^{f(x)} (4x^2 yy^2)dydx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01st Quadrant restriction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[= \int_a^b 2x^2f(x)^2  \frac{f(x)^3}{3}  2x^2h(x)^2 +\frac{h(x)^3}{3} dx\] \[= \int_a^b[ 2x^2(f(x)  h(x))(f(x) + h(x))  \frac{1}{3}(f(x)^3  h(x)^3) ]dx\] Can't do much more without knowing f(x) and h(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the bounds of y=cbrt(x) and y=x^3 are 1 and 0 so I believe a=0, b=1; h(x)=0; f(x)=1 prolly should a used c and d for those eh...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are your bounds on x? Or what is the domain you're integrating the function over?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is the domain? or are you trying to find the area?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the shaded area is the D part, im assuming that meant Domain in the book :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, ok. So you are integrating over that domain.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you should set it up as either \[\int\limits_{0}^{1}\int\limits_{x^3}^{\sqrt[3]{x}}{4x^2 y y^2}\ dydx\] or \[\int\limits_{0}^{1}\int\limits_{\sqrt[3]{y}}^{y^3}{4x^2 y y^2}\ dxdy\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is that [sqrt(y),y] or [sqrt(y),y^3]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i saw them changing stuff in the book, but at 3am it was hard to stay focuesd lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The limits on x are x=cubedRoot(y) to x=cubed(y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can do either version
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