Volumes with Cross Sections: Triangles/Semicircles Practice Test

$\displaystyle \int_{0}^{9} \frac{\sqrt3}{4},(2\sqrt{9-y})^2,dy$

$\displaystyle \int_{0}^{9} \frac12,(2\sqrt{9-y})^2,dy$

$\displaystyle \int_{-3}^{3} \frac{\sqrt3}{4},(9-x^2)^2,dx$

$\displaystyle \int_{0}^{9} \frac{\pi}{8},(2\sqrt{9-y})^2,dy$

$\displaystyle \int_{0}^{9} \frac{\sqrt3}{16},(2\sqrt{9-y})^2,dy$