### Video Transcript

The base of a solid is the first quadrant region bounded by π¦ equals the fourth root of four minus π₯ squared. Each cross section perpendicular to the π₯-axis is a square with one edge in the π₯π¦-plane. What is the volume of the solid?

Weβll find the volume of the solid bounded by π¦ equals the fourth root of four minus π₯ squared by integrating the area of the cross sections perpendicular to and bounded by that function. In our case, each cross section perpendicular to the π₯-axis is a square with one edge in the π₯π¦-plane. To integrate the areas of the cross sections, we need to know the limits of integration. And we can find these by looking at the boundary function π¦ equals the fourth root of four minus π₯ squared.

This function crosses the π₯-axis when π¦ is equal to zero. That is, when zero is the fourth root of four minus π₯ squared. Taking the fourth power on both sides, we have zero equals four minus π₯ squared. And that gives us π₯ equals plus or minus two. To sketch the function, letβs find the value of π¦ when π₯ is equal to zero. When π₯ is equal to zero, π¦ equals the fourth root of four minus zero, which is the fourth root of four. The fourth root of four is the square root of two. Since the base of our solid is in the first quadrant, weβll take the positive square root of two.

The base of our solid is the first quadrant region bounded by this function. Our solid also has cross sections perpendicular to the π₯-axis with one edge in the π₯π¦-plane. The sides of each cross-sectional square have length π¦ so that the area of each square is π¦ squared. But remember, π¦ is equal to the fourth root of four minus π₯ squared. So π¦ squared is the square of this. Remember, though, that the fourth root of four minus π₯ squared is four minus π₯ squared to the power one over four.

So the fourth root of four minus π₯ squared all squared is equal to four minus π₯ squared to the power two over four, which is actually the square root of four minus π₯ squared. So that π¦ squared is the square root of four minus π₯ squared. And this is the area of each square, depending on the value of π₯.

The volume of the solid is the integral of the area. And since the base of the solid is the first quadrant region bounded by the function π¦ equals the fourth root of four minus π₯ squared, the limits of the integration are zero and two. Weβve just found that the area is the square root of four minus π₯ squared. So the volume is the integral from zero to two of the square root of four minus π₯ squared with respect to π₯. To integrate this, we use trigonometric substitution.

Letting π₯ equal two sin π’, then dπ₯ is two cos π’ dπ’. And the square root of four minus π₯ squared is equal to the square root of four minus four sin squared π’. We have a common factor of four, which we can take the square root of and take outside. This gives us two times the square root of one minus sin squared π’. We can use the trigonometric identity cos squared π’ is one minus sin squared π’ so that the square root of four minus π₯ squared is equal to two times cos π’.

Now, letβs substitute dπ₯ equals two cos π’ dπ’ and the square root of four minus π₯ squared equals two cos π’ into our integral. And we have the volume is the integral of two cos π’ times two cos π’ with respect to π’. And thatβs equal to the integral of four cos squared π’ dπ’. Remember though that the volume is a definite integral. And since weβve made a substitution for π₯, weβll need to change the limits of integration. We need to find the value of two sin π’ when π₯ is zero and when π₯ is two.

When π₯ is zero, two sin π’ is zero means that sin π’ is equal to zero. And within our boundaries, this is true for π’ equal to zero. Our lower bound is, therefore, zero. For the upper bound, when π₯ is equal to two, we have two sin π’ equal to two. Dividing both sides by two, we have sin π’ equal to one. And the value of π’ within our boundaries for sin π’ equal to one is π’ equal to π by two. Our new upper limit with the substitution π₯ equals two sin π’ is π by two. The volume of our solid is then the integral from zero to π by two of four cos squared π’ dπ’.

Now, as four is a scaler, we can take this outside the integral. And we have four times the integral between zero and π by two of cos squared π’ dπ’. The standard formula for the integral of cos squared ππ₯ dπ₯ is π₯ over two plus sin two ππ₯ over four π. In our case, π is equal to one. So we have a volume equal to four times π’ over two plus sin two π’ over four evaluated between zero and π by two. Substituting π’ equals π by two, we have the volume equal to four times π by two over two plus sin π over four minus β with a substitution π’ equal to zero β zero over two plus sin zero over four. π by two divided by two is equal to π by four. sin π is equal to zero. So sin π over four is also equal to zero. Zero over two is equal to zero. And sin zero over four is equal to zero. So weβre left with four times π by four. We can cancel the fours.

So we have the volume of the solid equals π.