Double Integration

**adam_leeds** · May 2nd 2011, 09:32 AM

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 $\frac{{a}^{ 2} }{3 }$

**TheEmptySet** · May 2nd 2011, 09:45 AM

Originally Posted by adam_leeds

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

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

**TKHunny** · May 2nd 2011, 09:48 AM

Where did you get y = 0? Why not use the first equation?

We probably should establish, due ot the first constraint, that a >= 0.

**adam_leeds** · May 7th 2011, 10:32 AM

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

**TheEmptySet** · May 7th 2011, 10:48 AM

Originally Posted by adam_leeds

That doesnt get a^2/3

ummm no.

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

This gives

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

$\int_{0}^{a} -2x+2\sqrt{a}\sqrt{x}dx=-x^2+\frac{4}{3}\sqrt{a}x^{\frac{3}{2}}\bigg|_{0}^{ a}=\frac{a^2}{3}$

**adam_leeds** · May 7th 2011, 10:52 AM

Missed a negative, thanks.

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