Question

Sketch the region enclosed by x+y2=12x+y2=12 and x+y=0x+y=0. Decide whether to integrate with respect to xx or yy, and then find the area of the region.

Answer 1

Answer:

$A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}$

Step-by-step explanation:

For this case we have these two functions:

$x+y^2 = 12$   (1)

$x+y=0$   (2)

And as we can see we have the figure attached.

For this case we select the x axis in order to calculate the area.

If we solve y from equation (1) and (2) we got:

$y = \pm \sqrt{12-x}$

$y = -x$

Now we can solve for the intersection points:

$\sqrt{12-x} = -\sqrt{12-x}$

$12-x = -12+x$

$2x=24 , x=12$

$\sqrt{12-x} =-x$

$12-x = x^2$

$x^2 +x -12=0$

$(x+4)*(x-3) =0$

And the solutions are $x =-4, x=3$

So then we have in total 3 intersection point $x=12, x=-4, x=3$

And we can find the area between the two curves separating the total area like this:

$\int_{-4}^3 |\sqrt{12-x} - (-x)| dx +\int_{3}^{12}|-\sqrt{12-x} -\sqrt{12-x}|dx$

$\int_{-4}^3 |\sqrt{12-x} + x| dx +\int_{3}^{12}|-2\sqrt{12-x}|dx$

We can separate the integrals like this:

$\int_{-4}^3 |\sqrt{12-x} dx +\int_{-4}^3 x +2\int_{3}^{12}\sqrt{12-x} dx$

For this integral $\int_{-4}^3 |\sqrt{12-x} dx$ we can use the u substitution with $u = 12-x$ and after apply and solve the integral we got:

$\int_{-4}^3 |\sqrt{12-x} dx =\frac{74}{3}$

The other integral:

$\int_{-4}^3 x dx = \frac{3^2 -(-4)^2}{2} =-\frac{7}{2}$

And for the other integral:

$2\int_{3}^{12}\sqrt{12-x} dx$

We can use the same substitution $u = 12-x$ and after replace and solve the integral we got:

$2\int_{3}^{12}\sqrt{12-x} dx =36$

So then the final area would be given adding the 3 results as following:

$A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}$