# Double Integration

• May 2nd 2011, 09:32 AM
Double Integration
Use double integration to calculate the area that liew between the curves

$\sqrt{x} + \sqrt{y} = \sqrt{a}$ $and x + y = a, a > 0$This is what i have done , between the limits on the y = a-x and y = 0 and for x = a and 0 $\iint dy dx \int a- x dx = ax - \frac{{x}^{ 2} }{2 } = {a}^{ 2} - \frac{{a}^{ 2} }{2 } = \frac{{a}^{ 2} }{2 }$ But according to my text book the answer is $\frac{{a}^{ 2} }{3 }$
• May 2nd 2011, 09:45 AM
TheEmptySet
Quote:

Use double integration to calculate the area that liew between the curves

$
\sqrt{x} + \sqrt{y} = \sqrt{a}
$

$
and x + y = a, a > 0
$

This is what i have done , between the limits on the y = a-x and y = 0 and for x = a and 0

$
\iint dy dx

\int a- x dx

= ax - \frac{{x}^{ 2} }{2 } = {a}^{ 2} - \frac{{a}^{ 2} }{2 } = \frac{{a}^{ 2} }{2 }
$

But according to my text book the answer is $\frac{{a}^{ 2} }{3 }
$

Your limits of integration are incorrect.

If you sketch the region you will see that the line

$y=a-x$

lies above the curves

$y=(\sqrt{a}-\sqrt{x})^2$

and they both have x intercept

$(a,0)$

This gives

$\int_{0}^{a}\int_{(\sqrt{a}-\sqrt{x})^2}^{a-x}1 dydx$
• May 2nd 2011, 09:48 AM
TKHunny
Where did you get y = 0? Why not use the first equation?

We probably should establish, due ot the first constraint, that a >= 0.
• May 7th 2011, 10:32 AM
Quote:

Originally Posted by TheEmptySet
This gives

$\int_{0}^{a}\int_{(\sqrt{a}-\sqrt{x})^2}^{a-x}1 dydx$

That doesnt get a^2/3
• May 7th 2011, 10:48 AM
TheEmptySet
Quote:

That doesnt get a^2/3

ummm no.

$(\sqrt{a}-\sqrt{x})^2=a-2\sqrt{ax}+x$

This gives

$\int_{0}^{a} \int_{a-2\sqrt{ax}+x}^{a-x}dydx =\int_{0}^{a}(a-x)-(a-2\sqrt{ax}+x)dx$

$\int_{0}^{a} -2x+2\sqrt{a}\sqrt{x}dx=-x^2+\frac{4}{3}\sqrt{a}x^{\frac{3}{2}}\bigg|_{0}^{ a}=\frac{a^2}{3}$
• May 7th 2011, 10:52 AM