# Area of a region

• May 5th 2009, 05:40 PM
ZosoPage
Area of a region
• May 5th 2009, 06:03 PM
mr fantastic
Quote:

Originally Posted by ZosoPage

• May 6th 2009, 10:51 AM
ZosoPage
Sorry again.

Here's the problem without any red x's.

Find the exact area of the region bounded by $\displaystyle y=\frac{x^3}{\sqrt{4-x^2}}, y=0, x=0, x=\sqrt{2}$
• May 6th 2009, 11:05 AM
skeeter
Quote:

Originally Posted by ZosoPage
Sorry again.

Here's the problem without any red x's.

Find the exact area of the region bounded by $\displaystyle y=\frac{x^3}{\sqrt{4-x^2}}, y=0, x=0, x=\sqrt{2}$

try the substitution ...

$\displaystyle u^2 = 4 - x^2$

you'll end up with an easier definite integral ...

$\displaystyle \int_{\sqrt{2}}^2 4 - u^2 \, du$
• May 6th 2009, 12:19 PM
ZosoPage
Quote:

Originally Posted by skeeter
try the substitution ...

$\displaystyle u^2 = 4 - x^2$

you'll end up with an easier definite integral ...

$\displaystyle \int_{\sqrt{2}}^2 4 - u^2 \, du$

I'm confused. How did you use that substitution to come up with that integral? I need to take this integral but I don't know how:

$\displaystyle \int_{0}^{\sqrt{2}}\frac{x^3}{\sqrt{4-x^2}}dx$
• May 6th 2009, 12:39 PM
skeeter
Quote:

Originally Posted by ZosoPage
I'm confused. How did you use that substitution to come up with that integral? I need to take this integral but I don't know how:

$\displaystyle \int_{0}^{\sqrt{2}}\frac{x^3}{\sqrt{4-x^2}}dx$

$\displaystyle u^2 = 4-x^2$

$\displaystyle x^2 = 4-u^2$

$\displaystyle 2x \, dx = -2u \, du$

$\displaystyle x \, dx = -u \, du$

$\displaystyle \int \frac{x^3}{\sqrt{4-x^2}} \, dx = \int \frac{x^2}{\sqrt{4-x^2}} \cdot x \, dx$

substitute ...

$\displaystyle \int \frac{4-u^2}{u} \cdot (-u) \, du$

$\displaystyle -\int 4 - u^2 \, du$

lower limit ... $\displaystyle x = 0$ , $\displaystyle u = 2$

upper limit ... $\displaystyle x = \sqrt{2}$ , $\displaystyle u = \sqrt{2}$

$\displaystyle -\int_2^{\sqrt{2}} 4 - u^2 \, du = \int_{\sqrt{2}}^2 4 - u^2 \, du$