using shells of thickness dx, we have
v = ∫[0,1] 2πrh dx
where r=1-x and h=y
= ∫[0,1] π(1-x)^2*e^x dx = (2e-5)π
Find the volume generated by revolving the area under the curve𝑦 = 𝑒𝑥, from 𝑥 = 0 to 𝑥 =1,
about the line 𝑥 = 1.
1 answer