Answer to Question #217608 in Calculus for Gayathri Pranathi

Question

Evaluate integration integration X(y-1) dA where R is the region bounded by y=1-x^2 and y=x^2-3

Answer 1

$\iint x(y-1)dR,$ where $R=\{y=1-x^2, y= x^2-3\}$

First find intersepts points of $y = 1-x^2$ and $y = x^2-3$:

$1-x^2 = x^2-3\\ 2x^2 = 4\\ x^2 = 2\\ x = \pm \sqrt2$

So $\iint x(y-1)dR = \int_{-\sqrt2}^{\sqrt 2}dx \int_{x^2-3}^{1-x^2}x(y-1)dy =$

$= \int_{-\sqrt2}^{\sqrt 2}xdx(\cfrac{y^2}{2}-y|_{x^2-3}^{1-x^2}) = \\ =\int_{-\sqrt2}^{\sqrt 2}x(\cfrac{(1-x^2)^2}{2}-(1-x^2)-(\cfrac{(x^2-3)^2}{2}-(x^2-3)))dx=\\ =\int_{-\sqrt2}^{\sqrt 2}x(\cfrac{4x^2-8}{2}-4+2x^2)dx = \\ = \int_{-\sqrt2}^{\sqrt 2}x(4x^2-8)dx = \int_{-\sqrt2}^{\sqrt 2}(4x^3-8x)dx = x^4 - 4x^2 |_{-\sqrt 2}^{\sqrt 2} =\\ =4 - 4*2 -(4-4*2) = 0$