Question

Sketch the region determined by the iterated integral and then evaluate the integral. For some of these integrals, it may be helpful to reverse the order of integration.

${\int }_0^3{\int }_0^{\sqrt{9-y^2}}\sqrt{9-x^2}dxdy$

Short answer

The value of integral is$18$.

Step-by-step solution

Given Information

Iterated integral is${\int }_0^3{\int }_0^{\sqrt{9-y^2}}\sqrt{9-x^2}dxdy$

Calculation of Graph

The graph is as below:

Simplification

Changing the order of integral

${\int }_0^3{\int }_0^{\sqrt{9-y^2}}\sqrt{9-x^2}dxdy={\int }_0^3{\int }_0^{\sqrt{9-x^2}}\sqrt{9-x^2}dydx$

Therefore

${\int }_0^3{\int }_0^{\sqrt{9-x^2}}\sqrt{9-x^2}dydx={\int }_0^3[y{]}_0^{\sqrt{9-x^2}}\sqrt{9-x^2}dx$

$={\int }_0^3\left(9-x^2\right)dx$

$\Rightarrow {\int }_0^3{\int }_0^{\sqrt{9-x^2}}\sqrt{9-x^2}dydx={\left[9x-\frac{x^3}{3}\right]}_0^3$

$18$