Evaluate the double integral ∬_R (x² + y²) dA, where R is the region bounded by the curves y = x² and y = 4 in the first quadrant. First, set up and evaluate the integral in the or... Evaluate the double integral ∬_R (x² + y²) dA, where R is the region bounded by the curves y = x² and y = 4 in the first quadrant. First, set up and evaluate the integral in the order dy dx, then change the order of integration to dx dy and evaluate the same. Read more

Steps to Solve

1. Understand the Region of Integration

The region ( R ) is bounded by the curves ( y = x^2 ) and ( y = 4 ). In the first quadrant, the intersection occurs when ( x^2 = 4 ), giving ( x = 2 ). Thus, the limits for ( y ) are from ( x^2 ) to ( 4 ) and for ( x ) from ( 0 ) to ( 2 ).

2. Set Up the Integral in the Order ( dy , dx )

The double integral is written as: $$ \int_{0}^{2} \int_{x^2}^{4} (x^2 + y^2) , dy , dx $$

3. Evaluate the Inner Integral

Evaluate ( \int_{x^2}^{4} (x^2 + y^2) , dy ): $$ = \left[ x^2y + \frac{y^3}{3} \right]_{x^2}^{4} $$

Calculating this: $$ = \left( x^2(4) + \frac{4^3}{3} \right) - \left( x^2(x^2) + \frac{(x^2)^3}{3} \right) $$ $$ = 4x^2 + \frac{64}{3} - \left( x^4 + \frac{x^6}{3} \right) $$

4. Evaluate the Outer Integral

Now integrate with respect to ( x ): $$ \int_{0}^{2} \left( 4x^2 + \frac{64}{3} - x^4 - \frac{x^6}{3} \right) , dx $$

Compute this integral step by step:

• ( \int 4x^2 , dx = \frac{4}{3} x^3 )
• ( \int \frac{64}{3} , dx = \frac{64}{3} x )
• ( \int x^4 , dx = \frac{1}{5} x^5 )
• ( \int \frac{x^6}{3} , dx = \frac{1}{21} x^7 )

So we have: $$ \left[ \frac{4}{3} x^3 + \frac{64}{3} x - \frac{1}{5} x^5 - \frac{1}{21} x^7 \right]_{0}^{2} $$ Calculating this gives the result for the first integral.

5. Set Up the Integral in the Order ( dx , dy )

Now we switch the order of integration. The limits for ( x ) will vary from ( 0 ) to ( \sqrt{y} ), and for ( y ), from ( 0 ) to ( 4 ): $$ \int_{0}^{4} \int_{0}^{\sqrt{y}} (x^2 + y^2) , dx , dy $$

6. Evaluate the Inner Integral Again

Integrate ( \int_{0}^{\sqrt{y}} (x^2 + y^2) , dx ): $$ = \left[ \frac{x^3}{3} + y^2 x \right]_{0}^{\sqrt{y}} $$ Calculate this to get: $$ = \left( \frac{(\sqrt{y})^3}{3} + y^2(\sqrt{y}) \right) = \left( \frac{y^{3/2}}{3} + y^{5/2} \right) $$

7. Evaluate the Outer Integral Again

Now, integrate this with respect to ( y ): $$ \int_{0}^{4} \left( \frac{y^{3/2}}{3} + y^{5/2} \right) , dy $$ Again, calculate this integral step by step to determine the final result.