# Evaluate the integral: \int_0^2 \int_0^x \sqrt {4-x^2} dy \ dx

## Question:

Evaluate the integral: {eq}\int_0^2 \int_0^x \sqrt {4-x^2} \ dy \ dx {/eq}

## Double Integral:

Here, we have to evaluate the given double integral. First, we integrate the given function with respect to one variable while the other variable held constant. Then we integrate the function with respect to second variable.

To evaluate the integral, we 'll substitution method. The substitution method converts the integral into a new integral by applying suitable substitution that is easy to solve.

We have to evaluate the integral: {eq}\int_0^2 \int_0^x \sqrt{4-x^2} \ dy \ dx {/eq}

{eq}\begin{align} \int_0^2 \int_0^x \sqrt {4-x^2} \ dy \ dx &=\int_0^2 \sqrt {4-x^2} \left (\int_0^x \ dy \right ) \ dx\ & \left [\text{Integrate the given function with respect to } y \right ]\\ &=\int_0^2 \sqrt {4-x^2} \left ( y \right )_0^x dx \ \left [ \int dx=x \right ]\\ &=\int_0^2 \sqrt {4-x^2}(x-0)dx\\ &=\int_0^2 x\sqrt {4-x^2}dx \end{align} {/eq}

To integrate the function with respect to {eq}x {/eq}, we use the substitution {eq}4-x^2=u {/eq}.

Differentiate both sides with respect to {eq}x {/eq}

{eq}\begin{align} 0-2x &=\dfrac{du}{dx}\\ xdx &=-\dfrac{du}{2} \end{align} {/eq}

As we are dealing with definite integral, we also need to update the limits of integration according to the substitution.

When

{eq}x=0 \Rightarrow t=4-0^2=4\\ x=2 \Rightarrow t=4-2^2=0 {/eq}

Applying these to the integral, we have:

{eq}\begin{align} \int_0^2 x\sqrt {4-x^2}dx &=\int_4^0 \sqrt{u}\left ( \dfrac{du}{-2} \right )\\ &=\dfrac{-1}{2}\int_4^0 \sqrt{u} du\\ &=\dfrac{-1}{2}\left ( \frac{u^{\frac{1}{2}+1}}{\dfrac{1}{2}+1} \right )_4^0\\ &=\dfrac{-1}{2}\left ( \frac{u^{\frac{3}{2}}}{\dfrac{3}{2}} \right )_4^0\\ &=-\dfrac{1}{3}\left ( 0-4^{\frac{3}{2}} \right )\\ &=\dfrac{1}{3}(2^3)\\ &=\dfrac{8}{3} \end{align} {/eq}

{eq}\color{blue}{\boxed{\int_0^2 \int_0^x \sqrt {4-x^2} \ dy \ dx=\dfrac{8}{3}}} {/eq}