Use a triple integral to compute the exact volume of the solid enclosed by 𝑦=2𝑥^2, 𝑦=1, 𝑥=0, 𝑧=0, and 𝑧=8−𝑦 where one corner of the bounded region is (0,0,8).

Question

Use a triple integral to compute the exact volume of the solid enclosed by y=2x², y=1, x=0, z=0, and z=8-y where one corner of the bounded region is (0,0,8).

Answer 1

The key thing to note here is that, if 2x^2 <=1, x <= √2/2

Then, using I[a,b] for the integral from a to b and E[a,b] for the evaluation from a to b,

I[0, √2/2] I [2x^2, 1] I[0, 8 - y] dz dy dx =

I[0, √2/2] I [2x^2, 1] z E[0, 8 - y] dy dx =

I[0, √2/2] I [2x^2, 1] 8 - y dy dx =

I[0, √2/2] 8y - y^2/2 E[2x^2, 1] dx =

I[0, √2/2] 8 - 1/2 - (16x^2 - 2x^4) dx =

I[0, √2/2] 15/2 - 16x^2 + 2x^4 dx =

15/2 x - 16/3 x^3 + 2/5 x^5 E[0, √2/2] =

15√2/4 - 4√2/3 + √2/20 =

37√2/15