Fx-y is a continuous function defined ouver x-y- -0-1- -0-1- Given the two constraints- x-y2 and y-x2- the volume under fx-y is

The volume under f(x,y)=∫∫ f(x,y)dxdy In case of horizonta strip, x is varying nbsp; from x=y^2 to x=√(y) and y is varying from y=0 to 1, therefore, the volu ...

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Engineering Mathematics

Double integrals and Triple Integrals

fx,y is a con...

Question

$f (x, y)$ is a continuous function defined ouver $(x, y) ϵ [0, 1] \times [0, 1]$ . Given the two constraints, $x > y^{2}$ and $y > x^{2}$ , the volume under $f (x, y)$ is

\int_{y = 0}^{y = 1} \int_{x = y^{2}}^{x = \sqrt{y}} f (x, y) d x d y

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\int_{y = x^{2}}^{y =} \int_{x = y^{2}}^{x = 1} f (x, y) d x d y

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\int_{y = 0}^{y = 1} \int_{x = 0}^{x = 1} f (x, y) d x d y

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\int_{y = 0}^{y = \sqrt{x}} \int_{x = 0}^{x = \sqrt{y}} f (x, y) d x d y

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Solution

The correct option is A $\int_{y = 0}^{y = 1} \int_{x = y^{2}}^{x = \sqrt{y}} f (x, y) d x d y$

The volume under $f (x, y) = \int \int f (x, y) d x d y$
In case of horizonta strip, $x$ is varying from $x = y^{2}$ to $x = \sqrt{y}$ and $y$ is varying from $y = 0$ to $1$ , therefore, the volume will be
$= \int_{0}^{1} \int_{y^{2}}^{\sqrt{y}} f (x, y) d x d y$

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f(x,y) is a continuous function defined ouver (x,y)ϵ[0,1]×[0,1]. Given the two constraints, x>y2 and y>x2, the volume under f(x,y) is

The correct option is A ∫y=1y=0∫x=√yx=y2f(x,y)dxdy The volume under f(x,y)=∫∫f(x,y)dxdy In case of horizonta strip, x is varying from x=y2 to x=√y and y is varying from y=0 to 1, therefore, the volume will be =∫10∫√yy2f(x,y)dxdy

$f (x, y)$ is a continuous function defined ouver $(x, y) ϵ [0, 1] \times [0, 1]$ . Given the two constraints, $x > y^{2}$ and $y > x^{2}$ , the volume under $f (x, y)$ is