Using the washer method, we have an outer radius of and inner radius . You have the right interval, so

Is this how you started? I get a slightly different answer.

On so your outer radius is and your inner radius . So you haveb)find the volume of the solid created by revolving M around y=7

113.964

Again, I get a different answer.

Which part don't you understand? The region is the base of the solid; the solid will be "sticking out" of the plane (above the xy-plane, that is). If you cut the solid (with a cut perpendicular to the x-axis), the cross section is an isosceles triangle. This means that the base of each triangle has length .c)let region M be the base of a solid whose cross sections perpendicular to the x-axis are isosceles right triangles, find the volume of the solid.

This is the one I don't understand. Please help me setup this problem.

The formula you need is where is the area of the cross section at .

I'm not sure how the solid can have as a base and still have circular cross sections perpendicular to the x-axis. Semicircles, perhaps?d)let region M be the base of a solid whose cross sections perpendicular to the x-axis are circles, find the volume of the solid.

not sure how to do this one either.

I assume you mean region is the base; you haven't specified . Your interval is correct (approximately). Each semicircle has a base that extends from the top of the region to the bottom of the region. So the diameter of each is . Use the formula I gave above.#2Region Q is bounded by the y-axis, the line y=5 and the curve e(x)

Let region R be the base of a solid whose cross sections perpendicular to the x-axis are semi-circles, find the volume of the solid.

Is the interval for the integral 0 to 1.609? And then what do I do? Need help setting it up.

Not quite. is roughly the intersection of the two curves, but the region goes beyond that (look at a graph). You want to solve for the upper bound.#3 Region T is bounded by the x-axis and the graphs of and

This one is really confusing me. All I know is the interval 0 to 4.415.

I recommend using the washer method and integrating with respect to . In that case, your lower bound is and your upper bound is approximately Solve both equations for (in terms of ; you only need the positive root for the second one), and set up your integral.

For the parts involving revolution around horizontal axes, the shell method would be easiest, but you could do it with washers by splitting the interval at the intersection of the curves and using two integrals.