# intint_Asqrt(xy-y^2)dxdyquadquadA={(x,y)inRR^2; 0<=y<=3, x/10<=y<=x} Would anyone help me?

Apr 24, 2018

$162$

#### Explanation:

For A={(x,y)inRR^2; 0<=y<=3, x/10<=y<=x}

$\int {\int}_{A} \sqrt{x y - {y}^{2}} \mathrm{dx} \mathrm{dy} = {\int}_{y = 0}^{3} \left({\int}_{x = y}^{10 y} \sqrt{x y - {y}^{2}} \mathrm{dx}\right) \mathrm{dy}$
$q \quad = {\int}_{y = 0}^{3} {\left(\frac{2}{3 y} {\left(x y - {y}^{2}\right)}^{\frac{3}{2}}\right)}_{x = y}^{10 y} \mathrm{dy}$

$q \quad = {\int}_{y = 0}^{3} \left[\frac{2}{3 y} {\left(10 {y}^{2} - {y}^{2}\right)}^{\frac{3}{2}} - \frac{2}{3 y} {\left({y}^{2} - {y}^{2}\right)}^{\frac{3}{2}}\right] \mathrm{dy}$
$q \quad = \frac{2}{3} {\int}_{y = 0}^{3} 27 {y}^{2} \mathrm{dy} = \frac{2}{3} \times \frac{27}{3} \left({3}^{3} - {0}^{3}\right) = 162$

To get some geometric intuition into this, let us take a look at this graphically. The region $A$ is shown shaded below :

The double integral actually means "breaking up" this region into tiny pieces (like the darkened rectangle $\mathrm{dx} \setminus \mathrm{dy}$ in the picture below) and summing up the integrand times the area over all such pieces.

We can decide to either integrate over $x$ first, which essentially means adding up over all rectangular pieces at the same $y$ to get the contribution from the horizontal strip shown below :

The limits for this integration is decided by how much $x$ can vary at this fixed value of $x$. From the geometry, it is clear that $x$ varies from $y$ to $10 y$ - hence the limits.

We now integrate over $y$, This means summing up over all the horizontal strips, from $y = 0$ to $y = 3$ :

You could have decided to integrate over $y$ first , so that you would have vertical strips first, which are then added over.Let me leave this as an exercise!