Find the volume inside the curve $x^2+4z^2+8y=16$ and on the positive side of $xz$-plane.

Question

Find the volume inside the curve $x^2+4z^2+8y=16$ and on the positive side of $xz$-plane.

What will be the volume of this as I'm getting wrong answer?

As per me the volume is this.

Please help me as I'm unable to get it.

Answer 1

@player3236 my answer is in the image, please let me know where i went wrong, Thank you in advance $$v = \int_0^2 \int_0^\sqrt{16-4z^2} 1/8(16-x^2 +4z^2) \,dx\,dz$$

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$$v = \int_0^2 \int_0^\sqrt{16-4z^2} (2-x^2/8 +z^2/2) \,dx\,dz$$

My solution for the above question

Answer 2

Correct integral - $$v = 4\int_0^2 \int_0^\sqrt{16-4z^2} 1/8(16-x^2 +4z^2) \,dx\,dz$$

Correct Answer is $$ {8pi} $$

below is the correct solution of the given question -