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## Homework Statement

[tex]

\displaystyle\int\int\sqrt{4-x^2-y^2} dA

[/tex]

[tex]

R{(x,y)|x^2+y^2\leq4 .. 0\leq x}

[/tex]

## The Attempt at a Solution

So far i have:

[tex]

\displaystyle\int^{\pi}_{0}\int^{r}_{0}\sqrt{4-r^2} rdrd\theta

[/tex]

Solving i get:

[tex]

\displaystyle\int^{\pi}_{0}\frac{-1}{3}(4-r^2)^{\frac{3}{2}}+\frac{1}{3}(4)^{\frac{3}{2}}d\theta

[/tex]

Am i on the right track, and if so how do i integrate the third root term?

edit//

ok i forgot the radius goes from 0 to 2... if i sub 2 into r, i can integrate it. But just to check, am i still doing this correctly?

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