# Video: Solving a First-Order Separable Differential Equation

Use the slicing method to find the volume of the solid whose base is the region under the parabola π¦ = 4 β π₯Β² in the first quadrant and whose cross-sectional slices are squares perpendicular to the π₯-axis with one edge in the π₯π¦-plane.

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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 π¦ is equal to four minus π₯ squared in the first quadrant and whose cross-sectional slices are squares perpendicular to the π₯-axis with one edge in the π₯π¦-plane.

We want to find the volume of a solid whose base is the region under the parabola π¦ is equal to four minus π₯ squared. Weβre looking in the first quadrant. And where the solid has cross-sectional slices which are squares perpendicular to the π₯-axis with one edge in the π₯π¦-plane. The first thing we can do is draw our parabola π¦ is equal to four minus π₯ squared. And π¦ is equal to zero, our equation, gives us π₯ squared is equal to four. So, our parabola crosses the π₯-axis at π₯ is equal to negative two and two. When π₯ is equal to zero, π¦ is equal to four. So, our parabola intersects the π¦-axis at π¦ is equal to four.

The base of our solid is the region under the parabola in the first quadrant. The cross sections of our solid are squares which are perpendicular to the π₯-axis with one edge in the π₯π¦-plane. To find our volume, we integrate the areas of the cross section where the limits of integration are defined by our parabola, the lower limit is zero, and the upper limit is two.

We know that the area of a square with side length π is equal to π squared. In our case, the length of the sides of each square is π¦. And the value of π¦ depends on where the square meets the π₯-axis. The area of each square is, therefore, π¦ squared. And since π¦ is equal to four minus π₯ squared, we have the area of the cross section equal to four minus π₯ squared all squared. So, we have the area of the cross section as a function of π₯. Our volume is, then, the integral between zero and two of four minus π₯ squared all squared with respect to π₯.

To make the integral a little easier, we can expand the bracket. This gives us the integral between zero and two of 16 minus eight π₯ squared plus π₯ to the power of four with respect to π₯. The integral of 16 with respect to π₯ between zero and two is 16π₯ between zero and two. The integral of negative eight π₯ squared with respect to π₯ is negative eight times π₯ to the power of three divided by three between zero and two. And the integral of π₯ to the power of four is π₯ to the five divided by five evaluated between zero and two.

Evaluating this, we have 16 times two minus eight times eight over three plus two to the five over five minus zero. Evaluating this gives us 256 divided by 15. The volume of the solid whose base is the region under the parabola π¦ is equal to four minus π₯ squared in the first quadrant where the cross-sectional slices are squares perpendicular to the π₯-axis with one edge in the π₯π¦-plane is equal to 256 divided by 15 units cubed.