Video: Finding the Rectangle Dimensions That Would Yield the Maximum Area given Its Perimeter Using Differentiation

Use the slicing method to find the volume of the solid whose base is the region under the parabola 𝑦 = 9 βˆ’ π‘₯Β² and above the π‘₯-axis and where slices perpendicular to the 𝑦-axis are squares.

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Video Transcript

Use the slicing method to find the volume of the solid whose base is the region under the parabola 𝑦 is equal to nine minus π‘₯ squared and above the π‘₯-axis and where slices perpendicular to the 𝑦-axis are squares.

We’re asked to find the volume of a solid whose base is defined by the parabola 𝑦 is equal to nine minus π‘₯ squared and above the π‘₯-axis. The slices of the solid are perpendicular to the 𝑦-axis and are squares. Let’s first draw the base region of our solid. Our parabola is 𝑦 is equal to nine minus π‘₯ squared so that when 𝑦 is equal to zero, π‘₯ squared is equal to nine so that π‘₯ is equal to positive or negative three. When π‘₯ is equal to zero, 𝑦 is equal to nine. And this is where our function crosses the 𝑦-axis.

The region under this curve and above the π‘₯-axis forms the base of our solid. Our cross sections are squares. And these are perpendicular to the 𝑦-axis. The length of the sides of each square is two π‘₯. And the value of π‘₯ depends on where the square crosses the 𝑦-axis. The area of a cross-sectional square is, therefore, two π‘₯ times two π‘₯, which is four π‘₯ squared. The volume of our solid, however, is an integral along the 𝑦-axis between 𝑦 is zero and nine of the area of the cross-sectional squares as a function of 𝑦. So, we need to convert our area into a function of 𝑦.

We know that 𝑦 is equal to nine minus π‘₯ squared, which gives us π‘₯ squared equal to nine minus 𝑦. Our area as a function of π‘₯ is four π‘₯ squared so that four π‘₯ squared is equal to nine minus 𝑦 times four. In other words, four π‘₯ squared is equal to four times nine minus 𝑦 so that our area, in fact, is four times nine minus 𝑦. And we can now work out our volume, which is the integral between zero and nine of four times nine minus 𝑦 with respect to 𝑦.

We can take the four outside since this is a constant multiple. We know that the integral of nine with respect to 𝑦 is nine times 𝑦. And we know that the integral of negative 𝑦 with respect to 𝑦 is negative 𝑦 squared over two. So, then, our volume is four times nine 𝑦 minus 𝑦 squared over two evaluated between zero and nine. This is four times nine times nine minus nine squared over two minus zero. That gives us four times 81 minus 40.5, which is 162. The volume of the solid whose base is the region under the parabola 𝑦 is nine minus π‘₯ squared and above the π‘₯-axis where slices are perpendicular to the 𝑦-axis and are squares is, therefore, 162 units cubed.

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