Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle.
2x + y2 = 48, x = y

Find the area of the region.

Answer 1

Answer: A = 58

Step-by-step explanation: The sketched region enclosed by the curves and the approximating rectangle are shown in the attachment.

From the sketches, the area will be integrated with respect to y.

To calculate the integral, first determine the limits, which will be the points where both curves meet.

In respect to y:

$2x+y^{2} = 48$

$2x= 48- y^{2}$

$x= 24 - \frac{y^{2}}{2}$

Finding limits:

$y= 24 - \frac{y^{2}}{2}$

$24 - \frac{y^{2}}{2}-y=0$

Multiply by 2 to facilitate calculations:

$48 - y^{2}-2y=0$

Resolving quadratic equation:

$y=\frac{-2+\sqrt{2^{2}+192} }{2}$

y = 6 and y = -8

Then, integral to calculate area will be with limits -8<y<6:

$A = \int {24-\frac{y^{2}}{2}-y } \, dy$

$A = 24y - \frac{y^{3}}{6}-\frac{y^{2}}{2}$

$A = 24.6 - \frac{6^{3}}{6}-\frac{6^{2}}{2}-[24.(-8) - \frac{(-8)^{3}}{6}-\frac{(-8)^{2}}{2}]$

A = 58

The area of the enclosed region is 58 square units.