why the iterated integral is written in the above form. I thought x=rcos, and y=rsin, so after simple substitution, shouldn't it be like: 2-(rcos/3)-(2rsin)/3?

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B2. Consider the solid below the plane with equation x +2y+3z = 6 and above the unit disk in the xy-plane centred at the origin. Express the volume as a double integral and as an iterated integral in Cartesian coordinates. Then evaluated the iterated integral by converting to polar coordinates. Solution. We can rearrange the equation of the plane to obtain z = 2-3-2, which is the height of the solid at the point (x, y). So the solid can be described as {(x, y, z) | x² + y² ≤ 1 and 0 ≤ z ≤ 2 - - } and its volume is given by the double integral (²--)dA, where D = {(x,y) | x² + y² ≤ 1}. This double integral can be expressed as the following iterated integral in Cartesian coordinates. (2-3-4) dy dx 1-x² However, the iterated integral is easier to evaluate once we convert it to polar coordinates, which we do B2. Consider the solid below the plane with equation x +2y+3z = 6 and above the unit disk in the xy-plane centred at the origin. Express the volume as a double integral and as an iterated integral in Cartesian coordinates. Then evaluated the iterated integral by converting to polar coordinates. Solution. We can rearrange the equation of the plane to obtain z = 2-3-2, which is the height of the solid at the point (x, y). So the solid can be described as {(x, y, z) | x² + y² ≤ 1 and 0 ≤ z ≤ 2 - - } and its volume is given by the double integral (²--)dA, where D = {(x,y) | x² + y² ≤ 1}. This double integral can be expressed as the following iterated integral in Cartesian coordinates. (2-3-4) dy dx 1-x² However, the iterated integral is easier to evaluate once we convert it to polar coordinates, which we do

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2 Consider the solid below the plane with equation x2y3z6 and above the unit disk in the x... View the full answer