Question

Evaluate \[ \iint_D x^2y\,dx\,dy, \] where \[ D:\;x^2+y^2\le16. \]

• A. \(2\times\frac{4^5}{15}\)
• B. \(\pi\times\frac{4^5}{15}\)
• C. \(0\)
• D. \(\pi\times\frac{4^5}{25}\)

Hint:

Always check symmetry before integrating. An odd function integrated over a symmetric region often evaluates immediately to zero, saving considerable computation.

Answer 1

The Correct Option is C

Concept: Before evaluating a double integral, always examine the symmetry of the integrand and region. The region \[ x^2+y^2\le16 \] is a circle centered at the origin and is symmetric about both coordinate axes. The integrand \[ f(x,y)=x^2y \] is odd in \(y\) because \[ f(x,-y)=x^2(-y)=-f(x,y). \] Whenever an odd function is integrated over a region symmetric about the \(x\)-axis, the positive and negative contributions cancel.

Step 1: Check symmetry of the region.
The circular region \[ x^2+y^2\le16 \] contains the point \((x,y)\) whenever it contains \((x,-y)\). Hence the region is symmetric about the \(x\)-axis.

Step 2: Check symmetry of the integrand.
\[ f(x,y)=x^2y. \] Replacing \(y\) by \(-y\), \[ f(x,-y)=x^2(-y)=-x^2y. \] Thus \(f(x,y)\) is odd in \(y\).

Step 3: Apply the symmetry property.
For every positive contribution above the \(x\)-axis there exists an equal negative contribution below the \(x\)-axis. Therefore, \[ \iint_D x^2y\,dx\,dy=0. \] \[ \boxed{0} \]