The area of the region can be found by evaluating:
\[ A = \int _{-1}^{\answer {2}} \answer {y-1+\sqrt {3-y}} \d y +\int _{\answer {2}}^{\answer {3}} \answer {2\sqrt {3-y}} \d y \]
The region \(R\) is bounded by \(y=3-x^2\) and \(y=x+1\).
By integrating with respect to \(y\), how many integrals are needed to express the area
of \(R\)? \(\answer {2}\)
The base of a certain solid is the region \(R\). Cross sections through the solid taken
parallel to the \(y\)-axis are semicircles.
To express the volume of a solid using a definite integral, we should:
integrate with
respect to \(x\). integrate with respect to \(y\).
An integral that expresses the volume of the solid is:
\[ V= \int _{x=\answer {-2}}^{x=\answer {1}}\answer {\frac {\pi }{8}(2-x-x^2)^2} \d x \]