Question

Evaluate the double integral \(\iint\limits_D {xcosydA}\)D is bounded by\(y = 0,y = {x^2},x = 1\)

Short answer

The solution of the given integral can be: \(\iint\limits_D {x\cos ydA} = \frac{1}{2} - \frac{{\cos (1)}}{2}\)

Step-by-step solution

Step 1:Find the limit of the given integral

It’s easier to integrate\(\int {x\cos {x^2}dx} \) than\(\int {x\cos \sqrt x dx} \).

Expect integrating first on y to set up the easier problem.

For help setting up the integralgraph the region of integration and include a vertical line in middle.

Note that y runs from \(0\)to\({x^2}\)and that x runs from\(0\)to\(1\).

Apply the integration

\(\begin{array}{l}I = \int\limits_0^1 {\int\limits_0^{{x^2}} {x\cos ydydx} } \\I = \int\limits_0^1 {x\left( {\sin y} \right)_0^{{x^2}}dx} \\I = \int\limits_0^1 {x\left( {\sin {x^2} - 0} \right)dx} \\I = \int\limits_0^1 {x\left( {\sin {x^2}} \right)dx} \\I = - \frac{1}{2}\left( {\cos {x^2}} \right)_0^1\\I = - \frac{1}{2}\left( {\cos 1 - \cos 0} \right)\\I = - \frac{1}{2} - \frac{{\cos (1)}}{2}\end{array}\)

Therefore, The solution of the given double integral can be given as:

\(\iint\limits_D {x\cos ydA} = \frac{1}{2} - \frac{{\cos (1)}}{2}\)