Answer: [tex]\bold{(a)\quad \dfrac{32}{3}\qquad (b)\quad \dfrac{32}{3}}[/tex]
Step-by-step explanation:
(a) First, find the x-coordinates where the two equations cross
y = -1 and y = 3 - x²
-1 = 3 - x²
-4 = -x²
4 = x²
± 2 = x → These are the upper and lower limits of your integral
Then subtract the two equations and integrate with upper bound of x = 2 and lower bound of x = -2
[tex]\int_{-2}^{+2}[(3-x^2)-(-1)]dx\\\\\\=\int_{-2}^2(4-x^2)dx\\\\\\=4x-\dfrac{x^3}{3}\bigg|_{-2}^{+2}\\\\\\=\bigg(8-\dfrac{8}{3}\bigg)-\bigg(-8+\dfrac{8}{3}\bigg)\\\\\\=\large\boxed{\dfrac{32}{3}}[/tex]
(b) We know the upper and lower bounds of the y-axis as y = 3 and y = -1
Next, find the equation that we need to integrate by solving for x.
y = 3 - x²
x² + y = 3
x² = 3 - y
x [tex]=\pm\sqrt{3-y}\\[/tex]
[tex]\rightarrow \qquad x=\sqrt{3-y}\quad and \quad x=-\sqrt{3-y}[/tex]
Now, subtract the two equations and integrate with upper bound of y = 3 and lower bound of y = -1
[tex]\int_{-1}^{+3}[(\sqrt{3-y})-(-\sqrt{3-y})]dy\\\\\\=\int_{-1}^{+3}(2\sqrt{3-y})dy\\\\\\=\dfrac{-4\sqrt{(3-y)^3}}{3}\bigg|_{-1}^{+3}\\\\\\=\bigg(0\bigg)-\bigg(-\dfrac{32}{3}\bigg)\\\\\\=\large\boxed{\dfrac{32}{3}}[/tex]
