Question

Evaluate the double integral. $$ \int_{0}^{2} \int_{0}^{2}\left(6-x^{2}\right) d y d x $$

Short answer

The value of the double integral is 16.

Step-by-step solution

Evaluate the inner integral

Consider the inner integral first. That is, evaluate \(\int_{0}^{2}(6-x^{2}) dy\). Since this integral is with respect to \(y\), treat \(x\) as a constant. Therefore, the integral becomes \((6-x^{2}) * y\)

Apply the limits of integration for y

After integrating, apply the limits of \(y\), which are from 0 to 2. So, it becomes \((6-x^{2}) * y \biggr|_{0}^{2}\). Substituting the upper and lower limits of \(y\), we get \(2*(6-x^{2})\).

Evaluate the outer integral

Now, we want to integrate \(2*(6-x^{2})\) with respect to \(x\) from 0 to 2. The integral is then \(\int_{0}^{2} 2*(6-x^{2}) dx\).

Calculate the integral

Evaluate \(\int_{0}^{2} 2*(6-x^{2}) dx\), which yields \(2*[6x-\frac{x^{3}}{3}] \biggr|_{0}^{2}\).

Apply the limits of integration for x

Now apply the upper and lower limits of \(x\). After putting these values into the expression, subtract the lower limit from the upper limit to get the final answer, which results in 16.