I've been rusty integrals-wise and decided to try this problem out. I just need someone to corroborate:

Volume of the solid under the curve:
$$z=16xy+200$$bounded by:
$$y=x^2$$$$y=8-y^2$$Setting the bounds equal to each other yields the following region:
$$-2\ge x\ge2$$$$x^2\ge y\ge8-x^2$$We set up the double integral:
$$\int_{-2}^{2}\int_{x^2}^{8-x^2}16xy+200dydx$$$$\int_{-2}^{2}\left[8xy^2+200y\right]_{x^2}^{8-x^2}dx$$$$\int_{-2}^{2}8x^5-120x^3-400x^2+502x+1600dx$$$$\left[(4/3)x^6-30x^4-(400/3)x^3+251x^2+1600x\right]_{-2}^{2}=4,266.666$$

Question

I've been rusty integrals-wise and decided to try this problem out. I just need someone to corroborate:

Volume of the solid under the curve:
$$z=16xy+200$$bounded by:
$$y=x^2$$$$y=8-y^2$$Setting the bounds equal to each other yields the following region:
$$-2\ge x\ge2$$$$x^2\ge y\ge8-x^2$$We set up the double integral:
$$\int_{-2}^{2}\int_{x^2}^{8-x^2}16xy+200dydx$$$$\int_{-2}^{2}\left[8xy^2+200y\right]_{x^2}^{8-x^2}dx$$$$\int_{-2}^{2}8x^5-120x^3-400x^2+502x+1600dx$$$$\left[(4/3)x^6-30x^4-(400/3)x^3+251x^2+1600x\right]_{-2}^{2}=4,266.666$$

across · Answer

The inequalities of the regions are backwards :P

anonymous · Answer

Your work looks good.